Step 1: Substitution for Simplification

To integrate the function $I = \int \frac{2x}{(x^2+1)(x^2+3)} dx$, we observe that the derivative of $x^2$ is $2x$. Let us use the method of substitution.

Let $t = x^2$. [By the Method of Substitution]

Then, the differential $dt = 2x \, dx$.

The integral transforms into:

Step 2: Partial Fraction Decomposition

We express the integrand as a sum of partial fractions:

Multiplying both sides by $(t+1)(t+3)$, we obtain the identity:

To find the constants $A$ and $B$, we use the method of equating coefficients or substituting critical values: [By Partial Fraction Theorem]

• Set $t = -1$: $1 = A(-1+3) \implies 1 = 2A \implies A = \frac{1}{2}$
• Set $t = -3$: $1 = B(-3+1) \implies 1 = -2B \implies B = -\frac{1}{2}$

Step 3: Evaluating the Integral

Substitute the constants back into the integral form:

Using the standard integral $\int \frac{1}{x} dx = \ln|x| + C$:

Step 4: Re-substitution and Simplification

Substituting $t = x^2$ back into the result:

Since $x^2+1 > 0$ and $x^2+3 > 0$ for all real $x$, the absolute value bars are redundant.

Final Solution: \frac{1}{2} \ln\left( \frac{x^2+1}{x^2+3} \right) + C

Step 1: Application of the Algebraic Strategy

To integrate the function $I = \int \frac{1}{x(x^n+1)} \, dx$, we apply the suggested hint to facilitate substitution. Multiply the numerator and denominator by $x^{n-1}$:

$I = \int \frac{x^{n-1}}{x \cdot x^{n-1}(x^n+1)} \, dx = \int \frac{x^{n-1}}{x^n(x^n+1)} \, dx$

Step 2: Substitution and Differential Transformation

Let $t = x^n$. Then, differentiating both sides with respect to $x$ yields:

$\frac{dt}{dx} = nx^{n-1} \implies dt = nx^{n-1} \, dx \implies x^{n-1} \, dx = \frac{dt}{n}$

Substituting these into the integral expression:

$I = \int \frac{1}{t(t+1)} \cdot \frac{dt}{n} = \frac{1}{n} \int \frac{1}{t(t+1)} \, dt$

Step 3: Decomposition via Partial Fractions

We decompose the integrand $\frac{1}{t(t+1)}$ into partial fractions:

$\frac{1}{t(t+1)} = \frac{A}{t} + \frac{B}{t+1}$

$1 = A(t+1) + Bt$

• Setting $t = 0$: $1 = A(1) \implies A = 1$
• Setting $t = -1$: $1 = B(-1) \implies B = -1$

Step 4: Integration

Substituting the constants $A$ and $B$ back into the integral:

$I = \frac{1}{n} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) \, dt$

$I = \frac{1}{n} \left[ \ln|t| - \ln|t+1| \right] + C$

$I = \frac{1}{n} \ln \left| \frac{t}{t+1} \right| + C$

Step 5: Final Back-Substitution

Substitute $t = x^n$ back into the expression:

$I = \frac{1}{n} \ln \left| \frac{x^n}{x^n+1} \right| + C$

Final Solution: $\frac{1}{n} \ln \left| \frac{x^n}{x^n+1} \right| + C$

Step 1: Setting up the Partial Fraction Decomposition

To integrate the rational function $I = \int \frac{x}{(x-1)^2(x+2)} \, dx$, we express the integrand as a sum of partial fractions. Since the denominator contains a repeated linear factor $(x-1)^2$ and a distinct linear factor $(x+2)$, the decomposition takes the following form [Per the Theory of Partial Fractions for repeated linear factors]:

$$\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$$

Step 2: Determining the Constants

Multiply both sides of the equation by the common denominator $(x-1)^2(x+2)$ to clear the fractions:

$$x = A(x-1)(x+2) + B(x+2) + C(x-1)^2$$

To solve for $A, B,$ and $C$, we substitute convenient values for $x$:

• Set $x = 1$: $1 = A(0) + B(1+2) + C(0) \implies 1 = 3B \implies B = \frac{1}{3}$
• Set $x = -2$: $-2 = A(0) + B(0) + C(-2-1)^2 \implies -2 = 9C \implies C = -\frac{2}{9}$
• Equate coefficients of $x^2$: $0 = A + C \implies A = -C \implies A = \frac{2}{9}$

Step 3: Rewriting the Integral

Substitute the constants back into the integral expression:

$$I = \int \left[ \frac{2/9}{x-1} + \frac{1/3}{(x-1)^2} - \frac{2/9}{x+2} \right] \, dx$$

Apply the linearity property of integration [By the Linearity of the Integral Operator]:

$$I = \frac{2}{9} \int \frac{1}{x-1} \, dx + \frac{1}{3} \int (x-1)^{-2} \, dx - \frac{2}{9} \int \frac{1}{x+2} \, dx$$

Step 4: Evaluating the Components

Applying standard integration formulas $\int \frac{1}{u} du = \ln|u|$ and $\int u^n du = \frac{u^{n+1}}{n+1}$:

$$I = \frac{2}{9} \ln|x-1| + \frac{1}{3} \left( \frac{(x-1)^{-1}}{-1} \right) - \frac{2}{9} \ln|x+2| + K$$

Simplifying the algebraic terms:

$$I = \frac{2}{9} \ln|x-1| - \frac{1}{3(x-1)} - \frac{2}{9} \ln|x+2| + K$$

Step 5: Final Simplification

Grouping the logarithmic terms using the property $\ln(a) - \ln(b) = \ln(\frac{a}{b})$:

$$I = \frac{2}{9} \ln\left| \frac{x-1}{x+2} \right| - \frac{1}{3(x-1)} + K$$

Final Solution: \frac{2}{9} \ln\left| \frac{x-1}{x+2} \right| - \frac{1}{3(x-1)} + K

Step 1: Factorizing the Denominator

To perform partial fraction decomposition, the denominator $D(x) = x^3 - x^2 - x + 1$ must be factorized into linear or irreducible quadratic factors.

Grouping the terms: $D(x) = x^2(x - 1) - 1(x - 1)$

Factoring out $(x - 1)$: $D(x) = (x^2 - 1)(x - 1)$

Applying the difference of squares identity $a^2 - b^2 = (a-b)(a+b)$: $D(x) = (x - 1)(x + 1)(x - 1) = (x - 1)^2(x + 1)$

Step 2: Partial Fraction Decomposition Setup

Given the repeated linear factor $(x-1)^2$ and the distinct linear factor $(x+1)$, we express the integrand as:

$\frac{3x + 5}{(x - 1)^2(x + 1)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{x + 1}$ [By the Theory of Partial Fractions for repeated roots]

Step 3: Solving for Coefficients A, B, and C

Multiplying both sides by $(x - 1)^2(x + 1)$ gives the identity:

$3x + 5 = A(x - 1)(x + 1) + B(x + 1) + C(x - 1)^2$

$3x + 5 = A(x^2 - 1) + Bx + B + C(x^2 - 2x + 1)$

$3x + 5 = (A + C)x^2 + (B - 2C)x + (-A + B + C)$

Equating coefficients of like powers of $x$:

• $x^2: A + C = 0 \implies A = -C$
• $x^1: B - 2C = 3$
• $x^0: -A + B + C = 5$

Substitute $A = -C$ into the constant term equation: $-(-C) + B + C = 5 \implies B + 2C = 5$

Solving the system of linear equations:

1) $B - 2C = 3$

2) $B + 2C = 5$

Adding (1) and (2): $2B = 8 \implies B = 4$

Substituting $B = 4$ into (2): $4 + 2C = 5 \implies 2C = 1 \implies C = \frac{1}{2}$

Calculating $A$: $A = -C = -\frac{1}{2}$

Step 4: Integration

Substituting the coefficients back into the partial fraction expression:

$I = \int \left( \frac{-1/2}{x - 1} + \frac{4}{(x - 1)^2} + \frac{1/2}{x + 1} \right) dx$

$I = -\frac{1}{2} \int \frac{1}{x - 1} dx + 4 \int (x - 1)^{-2} dx + \frac{1}{2} \int \frac{1}{x + 1} dx$

Applying the standard integrals $\int \frac{1}{u} du = \ln|u| + C$ and $\int u^n du = \frac{u^{n+1}}{n+1} + C$:

$I = -\frac{1}{2} \ln|x - 1| + 4 \left( \frac{(x - 1)^{-1}}{-1} \right) + \frac{1}{2} \ln|x + 1| + C$

$I = \frac{1}{2} \ln \left| \frac{x + 1}{x - 1} \right| - \frac{4}{x - 1} + C$

Visual Representation of Decomposition Logic

Final Solution: $\frac{1}{2} \ln \left| \frac{x + 1}{x - 1} \right| - \frac{4}{x - 1} + C$

Step 1: Decomposition of the Rational Function

Given the integrand: $I = \int \frac{1}{x^4 - 1} \, dx$.

We factor the denominator using the difference of squares identity: $a^2 - b^2 = (a-b)(a+b)$.

$x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1)$ [Algebraic Factorization Theorem].

Step 2: Partial Fraction Decomposition

We express the integrand as a sum of partial fractions:

$\frac{1}{(x-1)(x+1)(x^2+1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx + D}{x^2 + 1}$

Multiplying by the common denominator $(x^4 - 1)$:

$1 = A(x+1)(x^2+1) + B(x-1)(x^2+1) + (Cx+D)(x-1)(x+1)$

Step 3: Solving for Coefficients

• Let $x = 1$: $1 = A(2)(2) \implies 4A = 1 \implies A = \frac{1}{4}$.
• Let $x = -1$: $1 = B(-2)(2) \implies -4B = 1 \implies B = -\frac{1}{4}$.
• Equating coefficients of $x^3$: $0 = A + B + C$. Since $A + B = 0$, $C = 0$.
• Let $x = 0$ (to find $D$): $1 = A(1)(1) + B(-1)(1) + D(-1)(1) \implies 1 = A - B - D$.
• Substituting $A = \frac{1}{4}$ and $B = -\frac{1}{4}$: $1 = \frac{1}{4} - (-\frac{1}{4}) - D \implies 1 = \frac{1}{2} - D \implies D = -\frac{1}{2}$.

Step 4: Integration of Partial Fractions

$I = \int \left( \frac{1/4}{x-1} - \frac{1/4}{x+1} - \frac{1/2}{x^2+1} \right) \, dx$

Applying the linearity property of integrals [Integral Linearity]:

$I = \frac{1}{4} \int \frac{1}{x-1} \, dx - \frac{1}{4} \int \frac{1}{x+1} \, dx - \frac{1}{2} \int \frac{1}{x^2+1} \, dx$

Using the standard integral formulas $\int \frac{1}{u} du = \ln|u|$ and $\int \frac{1}{x^2+1} dx = \tan^{-1}(x) + C$:

$I = \frac{1}{4} \ln|x-1| - \frac{1}{4} \ln|x+1| - \frac{1}{2} \tan^{-1}(x) + C$

Step 5: Logarithmic Simplification

Using the quotient rule for logarithms: $\ln(a) - \ln(b) = \ln(\frac{a}{b})$:

$I = \frac{1}{4} \ln \left| \frac{x-1}{x+1} \right| - \frac{1}{2} \tan^{-1}(x) + C$

Final Solution: $\frac{1}{4} \ln \left| \frac{x-1}{x+1} \right| - \frac{1}{2} \tan^{-1}(x) + C$

Step 1: Factorization and Decomposition Setup

The given integrand is $I = \int \frac{2x-3}{(x^2-1)(2x+3)} \, dx$. First, factorize the quadratic term $(x^2-1)$ as $(x-1)(x+1)$ [By difference of squares identity $a^2-b^2 = (a-b)(a+b)$].

The expression becomes: $\frac{2x-3}{(x-1)(x+1)(2x+3)}$.

We decompose this into partial fractions as follows:

$\frac{2x-3}{(x-1)(x+1)(2x+3)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{2x+3}$

Step 2: Determining Constants $A, B,$ and $C$

Multiplying both sides by the denominator $(x-1)(x+1)(2x+3)$, we obtain:

$2x-3 = A(x+1)(2x+3) + B(x-1)(2x+3) + C(x-1)(x+1)$

• For $A$ (Let $x = 1$): $2(1)-3 = A(1+1)(2(1)+3) \Rightarrow -1 = A(2)(5) \Rightarrow A = -\frac{1}{10}$
• For $B$ (Let $x = -1$): $2(-1)-3 = B(-1-1)(2(-1)+3) \Rightarrow -5 = B(-2)(1) \Rightarrow B = \frac{5}{2}$
• For $C$ (Let $x = -\frac{3}{2}$): $2(-\frac{3}{2})-3 = C(-\frac{3}{2}-1)(-\frac{3}{2}+1) \Rightarrow -6 = C(-\frac{5}{2})(-\frac{1}{2}) \Rightarrow -6 = C(\frac{5}{4}) \Rightarrow C = -\frac{24}{5}$

Step 3: Integration of Partial Fractions

Substituting the constants back into the integral expression:

$I = \int \left( -\frac{1}{10(x-1)} + \frac{5}{2(x+1)} - \frac{24}{5(2x+3)} \right) dx$

Applying the linearity property of integrals $\int (af(x) + bg(x)) dx = a \int f(x) dx + b \int g(x) dx$ and the standard integral $\int \frac{1}{ax+b} dx = \frac{1}{a} \ln|ax+b| + C$:

$I = -\frac{1}{10} \ln|x-1| + \frac{5}{2} \ln|x+1| - \frac{24}{5} \cdot \frac{1}{2} \ln|2x+3| + K$

$I = -\frac{1}{10} \ln|x-1| + \frac{5}{2} \ln|x+1| - \frac{12}{5} \ln|2x+3| + K$

Step 4: Geometric Representation of the Integrand

Final Solution: $-\frac{1}{10} \ln|x-1| + \frac{5}{2} \ln|x+1| - \frac{12}{5} \ln|2x+3| + K$

Step 1: Decomposition via Partial Fractions

To evaluate the integral $I = \int \frac{x}{(x-1)(x-2)} dx$, we express the integrand as a sum of partial fractions [Per the Method of Partial Fraction Decomposition].

Let the integrand be represented as:

$\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$

Multiplying both sides by the denominator $(x-1)(x-2)$, we obtain the identity:

$x = A(x-2) + B(x-1)$

Step 2: Determining Constants A and B

To find the values of constants $A$ and $B$, we apply the method of strategic substitution:

• Set $x = 1$: $1 = A(1-2) + B(1-1)$ $1 = -A \implies A = -1$
• Set $x = 2$: $2 = A(2-2) + B(2-1)$ $2 = B \implies B = 2$

Step 3: Integration of Partial Fractions

Substituting the values of $A$ and $B$ back into the integral expression:

$I = \int \left( \frac{-1}{x-1} + \frac{2}{x-2} \right) dx$

Applying the linearity property of integrals [Per the Additivity of Integration]:

$I = -\int \frac{1}{x-1} dx + 2 \int \frac{1}{x-2} dx$

Step 4: Applying Logarithmic Integration Rule

Using the fundamental integral identity $\int \frac{1}{u} du = \ln|u| + C$:

$I = -\ln|x-1| + 2\ln|x-2| + C$

Using the logarithmic property $n\ln(a) = \ln(a^n)$:

$I = \ln|x-2|^2 - \ln|x-1| + C$

Using the quotient rule $\ln(a) - \ln(b) = \ln(\frac{a}{b})$:

$I = \ln\left| \frac{(x-2)^2}{x-1} \right| + C$

Visual Representation of Function Asymptotes

Final Solution: $\ln\left| \frac{(x-2)^2}{x-1} \right| + C$

Step 1: Decomposition via Partial Fractions

To integrate the rational function $f(x) = \frac{x}{(x+1)(x+2)}$, we express the integrand as a sum of partial fractions [Per the Method of Partial Fraction Decomposition for distinct linear factors]. We set up the identity:

$\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$

Multiplying both sides by the denominator $(x+1)(x+2)$, we obtain the basic equation:

$x = A(x+2) + B(x+1)$

Step 2: Determining Constants A and B

To solve for the unknown coefficients, we evaluate the equation at the roots of the denominator factors [By the Principle of Undetermined Coefficients]:

• Setting $x = -1$:
  $-1 = A(-1+2) + B(-1+1)$
  $-1 = A(1) + 0 \implies A = -1$
• Setting $x = -2$:
  $-2 = A(-2+2) + B(-2+1)$
  $-2 = 0 + B(-1) \implies B = 2$

Step 3: Integration of the Decomposed Form

Substitute the constants back into the integral expression:

$\int \frac{x}{(x+1)(x+2)} \, dx = \int \left( \frac{-1}{x+1} + \frac{2}{x+2} \right) \, dx$

Applying the linearity property of integrals [Per the Sum/Difference Rule of Integration]:

$= -\int \frac{1}{x+1} \, dx + 2\int \frac{1}{x+2} \, dx$

Using the standard integral formula $\int \frac{1}{u} \, du = \ln|u| + C$:

$= -\ln|x+1| + 2\ln|x+2| + C$

Step 4: Algebraic Simplification

Utilizing logarithmic properties, specifically $n\ln(a) = \ln(a^n)$ and $\ln(a) - \ln(b) = \ln(\frac{a}{b})$:

$= \ln|x+2|^2 - \ln|x+1| + C$

$= \ln\left| \frac{(x+2)^2}{x+1} \right| + C$

Visual Representation of the Integrand Behavior

Final Solution: $-\ln|x+1| + 2\ln|x+2| + C$ or $\ln\left| \frac{(x+2)^2}{x+1} \right| + C$

Given Variables & Initial Setup

We are required to evaluate the integral of the rational function:

$I = \int \frac{1}{x^2 - 9} \, dx$

This integral takes the form $\int \frac{1}{x^2 - a^2} \, dx$, where $a = 3$. We shall employ the Method of Partial Fractions to decompose the integrand.

Step 1: Factorizing the Denominator

Using the algebraic identity for the difference of squares, $a^2 - b^2 = (a - b)(a + b)$, we factor the denominator:

$x^2 - 9 = (x - 3)(x + 3)$ [By Algebraic Identity $a^2 - b^2$]

Step 2: Partial Fraction Decomposition

We express the integrand as a sum of partial fractions:

$\frac{1}{(x - 3)(x + 3)} = \frac{A}{x - 3} + \frac{B}{x + 3}$

Multiplying both sides by $(x - 3)(x + 3)$ yields:

$1 = A(x + 3) + B(x - 3)$

Step 3: Solving for Coefficients A and B

To find $A$ and $B$, we equate the coefficients or substitute convenient values for $x$:

• Let $x = 3$: $1 = A(3 + 3) + B(0) \implies 6A = 1 \implies A = \frac{1}{6}$
• Let $x = -3$: $1 = A(0) + B(-3 - 3) \implies -6B = 1 \implies B = -\frac{1}{6}$

Step 4: Integration

Substituting the values of $A$ and $B$ back into the integral:

$I = \int \left( \frac{1/6}{x - 3} - \frac{1/6}{x + 3} \right) \, dx$

$I = \frac{1}{6} \int \frac{1}{x - 3} \, dx - \frac{1}{6} \int \frac{1}{x + 3} \, dx$

Using the standard integral formula $\int \frac{1}{u} \, du = \ln|u| + C$ [By Integration Principle of Linear Forms]:

$I = \frac{1}{6} \ln|x - 3| - \frac{1}{6} \ln|x + 3| + C$

Step 5: Applying Logarithmic Properties

Using the property $\ln(m) - \ln(n) = \ln\left(\frac{m}{n}\right)$:

$I = \frac{1}{6} \ln\left| \frac{x - 3}{x + 3} \right| + C$

Final Solution: $\frac{1}{6} \ln\left| \frac{x - 3}{x + 3} \right| + C$

Step 1: Decomposition of the Rational Function

To integrate $\int \frac{5x}{(x+1)(x^2-4)} dx$, we first factorize the quadratic term in the denominator. Recall the algebraic identity $a^2 - b^2 = (a-b)(a+b)$.

$x^2 - 4 = (x-2)(x+2)$

Therefore, the integrand becomes:

$\frac{5x}{(x+1)(x-2)(x+2)}$

[By Partial Fraction Decomposition theorem, since all factors are linear and distinct, we assume]:

$\frac{5x}{(x+1)(x-2)(x+2)} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+2}$

Step 2: Determining the Constants A, B, and C

Multiply both sides by $(x+1)(x-2)(x+2)$ to obtain the identity:

$5x = A(x-2)(x+2) + B(x+1)(x+2) + C(x+1)(x-2)$

$5x = A(x^2 - 4) + B(x^2 + 3x + 2) + C(x^2 - x - 2)$

• To find $A$, set $x = -1$: $5(-1) = A(-1-2)(-1+2) \implies -5 = A(-3)(1) \implies A = \frac{5}{3}$
• To find $B$, set $x = 2$: $5(2) = B(2+1)(2+2) \implies 10 = B(3)(4) \implies 10 = 12B \implies B = \frac{10}{12} = \frac{5}{6}$
• To find $C$, set $x = -2$: $5(-2) = C(-2+1)(-2-2) \implies -10 = C(-1)(-4) \implies -10 = 4C \implies C = -\frac{10}{4} = -\frac{5}{2}$

Step 3: Integration of the Partial Fractions

The integral can now be expressed as the sum of three elementary integrals:

$\int \frac{5x}{(x+1)(x^2-4)} dx = \frac{5}{3} \int \frac{1}{x+1} dx + \frac{5}{6} \int \frac{1}{x-2} dx - \frac{5}{2} \int \frac{1}{x+2} dx$

[Applying the standard integral rule $\int \frac{1}{x} dx = \ln|x| + C$]:

$= \frac{5}{3} \ln|x+1| + \frac{5}{6} \ln|x-2| - \frac{5}{2} \ln|x+2| + C$

Visual Representation of Partial Fraction Components

Step 4: Simplification

To provide a more compact form, we observe the coefficients $\frac{5}{3}, \frac{5}{6}, -\frac{5}{2}$. Finding a common denominator ($6$):

$= \frac{10}{6} \ln|x+1| + \frac{5}{6} \ln|x-2| - \frac{15}{6} \ln|x+2| + C$

$= \frac{5}{6} [2\ln|x+1| + \ln|x-2| - 3\ln|x+2|] + C$

[Using logarithmic power rule $n\ln(a) = \ln(a^n)$ and quotient rule $\ln(a) - \ln(b) = \ln(a/b)$]:

$= \frac{5}{6} \ln \left| \frac{(x+1)^2(x-2)}{(x+2)^3} \right| + C$

Final Solution: \frac{5}{3} \ln|x+1| + \frac{5}{6} \ln|x-2| - \frac{5}{2} \ln|x+2| + C

Step 1: Algebraic Simplification via Substitution

The integrand is $I = \int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)} \, dx$. Since the expression contains only even powers of $x$, let $u = x^2$. The integrand becomes a rational function of $u$:

$f(u) = \frac{(u+1)(u+2)}{(u+3)(u+4)} = \frac{u^2 + 3u + 2}{u^2 + 7u + 12}$

Since the degree of the numerator is equal to the degree of the denominator, we perform polynomial long division [Per Euclidean Division Theorem]:

$f(u) = \frac{(u^2 + 7u + 12) - 4u - 10}{u^2 + 7u + 12} = 1 - \frac{4u + 10}{(u+3)(u+4)}$

Step 2: Partial Fraction Decomposition

We decompose the fractional part into partial fractions [By Method of Undetermined Coefficients]:

$\frac{4u + 10}{(u+3)(u+4)} = \frac{A}{u+3} + \frac{B}{u+4}$

$4u + 10 = A(u+4) + B(u+3)$

• Setting $u = -3$: $4(-3) + 10 = A(-3+4) \implies -2 = A$
• Setting $u = -4$: $4(-4) + 10 = B(-4+3) \implies -6 = -B \implies B = 6$

Thus, $f(u) = 1 - \left( \frac{-2}{u+3} + \frac{6}{u+4} \right) = 1 + \frac{2}{u+3} - \frac{6}{u+4}$

Step 3: Back-Substitution and Integration

Substituting $u = x^2$ back into the expression:

$I = \int \left( 1 + \frac{2}{x^2 + 3} - \frac{6}{x^2 + 4} \right) \, dx$

$I = \int 1 \, dx + 2 \int \frac{1}{x^2 + (\sqrt{3})^2} \, dx - 6 \int \frac{1}{x^2 + 2^2} \, dx$

Applying the standard integration formula $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$:

$I = x + 2 \left( \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) \right) - 6 \left( \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) \right) + C$

Visual Representation of Integral Geometry

Step 4: Simplification of Final Result

$I = x + \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) - 3 \tan^{-1}\left(\frac{x}{2}\right) + C$

Final Solution: $x + \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) - 3 \tan^{-1}\left(\frac{x}{2}\right) + C$

Step 1: Substitution and Variable Transformation

Given the integral: $I = \int \frac{\cos x}{(1-\sin x)(2-\sin x)} dx$.

Following the hint, let $\sin x = t$. Differentiating both sides with respect to $x$:

$\frac{d}{dx}(\sin x) = \frac{dt}{dx} \implies \cos x \, dx = dt$ [By the Chain Rule of differentiation].

Substituting these into the integral, the expression transforms into a rational function in terms of $t$:

$I = \int \frac{1}{(1-t)(2-t)} dt$

Step 2: Partial Fraction Decomposition

To integrate $\frac{1}{(1-t)(2-t)}$, we decompose the integrand into partial fractions:

$\frac{1}{(1-t)(2-t)} = \frac{A}{1-t} + \frac{B}{2-t}$

Multiplying both sides by the denominator $(1-t)(2-t)$:

$1 = A(2-t) + B(1-t)$

To solve for $A$ and $B$, we use the method of strategic substitution:

• Set $t = 1$: $1 = A(2-1) + B(0) \implies A = 1$.
• Set $t = 2$: $1 = A(2-2) + B(1-2) \implies 1 = -B \implies B = -1$.

[By the Identity Principle for polynomials].

Step 3: Integration of the Partial Fractions

Substituting $A=1$ and $B=-1$ back into the integral expression:

$I = \int \left( \frac{1}{1-t} - \frac{1}{2-t} \right) dt$

Applying the standard integral formula $\int \frac{1}{ax+b} dx = \frac{1}{a} \ln|ax+b| + C$:

$I = \int \frac{1}{1-t} dt - \int \frac{1}{2-t} dt$

$I = \frac{\ln|1-t|}{-1} - \frac{\ln|2-t|}{-1} + C$ [Applying the Linear Substitution Rule].

$I = -\ln|1-t| + \ln|2-t| + C$

Step 4: Final Substitution and Simplification

Using the logarithmic property $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$:

$I = \ln\left| \frac{2-t}{1-t} \right| + C$

Substitute $t = \sin x$ back into the result to obtain the function in terms of $x$:

$I = \ln\left| \frac{2-\sin x}{1-\sin x} \right| + C$

Geometric Representation of the Integral Mapping

Final Solution: \ln\left| \frac{2-\sin x}{1-\sin x} \right| + C

Step 1: Identifying the Rational Function Type

The given integral is $I = \int \frac{1-x^2}{x(1-2x)} dx$. First, observe that the degree of the numerator ($2$) is equal to the degree of the denominator ($x(1-2x) = x - 2x^2$, which is also $2$). [Per the definition of partial fractions, we must perform algebraic division before applying decomposition if the degree of the numerator is $\geq$ degree of the denominator].

Step 2: Performing Algebraic Division

Expand the denominator: $x(1-2x) = -2x^2 + x$.

Rewrite the integrand: $\frac{-x^2 + 1}{-2x^2 + x}$.

Perform the division $\frac{-x^2 + 1}{-2x^2 + x} = \frac{1}{2} + \frac{-\frac{1}{2}x + 1}{-2x^2 + x}$.

Verification: $\frac{1}{2}(-2x^2 + x) + (-\frac{1}{2}x + 1) = -x^2 + \frac{1}{2}x - \frac{1}{2}x + 1 = -x^2 + 1$. [By the Division Algorithm for Polynomials].

Step 3: Partial Fraction Decomposition

We decompose the remainder term: $\frac{-\frac{1}{2}x + 1}{x(1-2x)} = \frac{A}{x} + \frac{B}{1-2x}$.

Equating numerators: $-\frac{1}{2}x + 1 = A(1-2x) + Bx$.

• Setting $x = 0$: $1 = A(1-0) \Rightarrow A = 1$.
• Setting $x = \frac{1}{2}$: $-\frac{1}{2}(\frac{1}{2}) + 1 = B(\frac{1}{2}) \Rightarrow -\frac{1}{4} + 1 = \frac{1}{2}B \Rightarrow \frac{3}{4} = \frac{1}{2}B \Rightarrow B = \frac{3}{2}$.

Step 4: Setting up the Integral

$I = \int \left( \frac{1}{2} + \frac{1}{x} + \frac{3/2}{1-2x} \right) dx$.

Applying the linearity property of integrals: $I = \frac{1}{2} \int 1 dx + \int \frac{1}{x} dx + \frac{3}{2} \int \frac{1}{1-2x} dx$.

Step 5: Integration and Final Computation

Using standard integrals $\int x^n dx = \frac{x^{n+1}}{n+1}$ and $\int \frac{1}{u} du = \ln|u|$:

$I = \frac{1}{2}x + \ln|x| + \frac{3}{2} \left( \frac{\ln|1-2x|}{-2} \right) + C$ [By substitution $u = 1-2x, du = -2dx$].

$I = \frac{x}{2} + \ln|x| - \frac{3}{4} \ln|1-2x| + C$.

Final Solution: \frac{x}{2} + \ln|x| - \frac{3}{4} \ln|1-2x| + C

Step 1: Decomposition via Partial Fractions

To integrate $I = \int \frac{1}{x(x^2+1)} \, dx$, we express the integrand as a sum of partial fractions. Since $x^2+1$ is an irreducible quadratic factor, the decomposition takes the following form:

$\frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx + C}{x^2+1}$ [By the Method of Partial Fractions for irreducible quadratic factors]

Step 2: Solving for Coefficients A, B, and C

Multiply both sides by the denominator $x(x^2+1)$:

$1 = A(x^2+1) + (Bx+C)x$

$1 = Ax^2 + A + Bx^2 + Cx$

$1 = (A+B)x^2 + Cx + A$

By comparing the coefficients of the powers of $x$ on both sides:

• Constant term: $A = 1$
• Coefficient of $x$: $C = 0$
• Coefficient of $x^2$: $A+B = 0 \implies 1+B = 0 \implies B = -1$

Step 3: Rewriting the Integral

Substitute the constants back into the partial fraction decomposition:

$I = \int \left( \frac{1}{x} + \frac{-x + 0}{x^2+1} \right) \, dx$

$I = \int \frac{1}{x} \, dx - \int \frac{x}{x^2+1} \, dx$

Step 4: Execution of Integration

For the first term, $\int \frac{1}{x} \, dx = \ln|x|$.

For the second term, use the method of substitution. Let $u = x^2 + 1$, then $du = 2x \, dx$, which implies $x \, dx = \frac{1}{2} du$.

$I = \ln|x| - \frac{1}{2} \int \frac{1}{u} \, du$ [By the power rule for integration of reciprocals]

$I = \ln|x| - \frac{1}{2} \ln|x^2+1| + C$

Step 5: Logarithmic Simplification

Using the property $a \ln(b) = \ln(b^a)$:

$I = \ln|x| - \ln(x^2+1)^{1/2} + C$

$I = \ln \left| \frac{x}{\sqrt{x^2+1}} \right| + C$

Final Solution: \log|x| - \frac{1}{2}\log(x^2+1) + C = \log \left| \frac{x}{\sqrt{x^2+1}} \right| + C

Step 1: Decomposition via Partial Fractions

The integrand is $f(x) = \frac{x}{(x^2+1)(x-1)}$. Since the denominator contains an irreducible quadratic factor $(x^2+1)$ and a linear factor $(x-1)$, we apply the method of partial fractions [Per the Partial Fraction Decomposition Theorem]. We express the integrand as:

$\frac{x}{(x^2+1)(x-1)} = \frac{Ax + B}{x^2 + 1} + \frac{C}{x - 1}$

Step 2: Solving for Coefficients

Multiplying both sides by the denominator $(x^2+1)(x-1)$, we obtain the identity:

$x = (Ax + B)(x - 1) + C(x^2 + 1)$

$x = Ax^2 - Ax + Bx - B + Cx^2 + C$

$x = (A + C)x^2 + (B - A)x + (C - B)$

Equating the coefficients of corresponding powers of $x$ on both sides:

• For $x^2$: $A + C = 0 \implies C = -A$
• For $x^1$: $B - A = 1$
• Constant: $C - B = 0 \implies C = B$

From $C = -A$ and $C = B$, we have $B = -A$. Substituting $B = -A$ into $B - A = 1$:

$-A - A = 1 \implies -2A = 1 \implies A = -\frac{1}{2}$

Consequently, $B = \frac{1}{2}$ and $C = \frac{1}{2}$.

Step 3: Setting up the Integral

The integral now takes the form:

$I = \int \left( \frac{-\frac{1}{2}x + \frac{1}{2}}{x^2 + 1} + \frac{\frac{1}{2}}{x - 1} \right) dx$

$I = -\frac{1}{2} \int \frac{x}{x^2 + 1} dx + \frac{1}{2} \int \frac{1}{x^2 + 1} dx + \frac{1}{2} \int \frac{1}{x - 1} dx$

Step 4: Evaluating the Components

Applying standard integration formulas [Table of Standard Integrals]:

• $\int \frac{x}{x^2 + 1} dx$: Let $u = x^2 + 1$, then $du = 2x dx$. Thus, $\frac{1}{2} \ln|x^2 + 1|$.
• $\int \frac{1}{x^2 + 1} dx = \tan^{-1}(x)$.
• $\int \frac{1}{x - 1} dx = \ln|x - 1|$.

$I = -\frac{1}{2} \cdot \left( \frac{1}{2} \ln|x^2 + 1| \right) + \frac{1}{2} \tan^{-1}(x) + \frac{1}{2} \ln|x - 1| + C$

$I = -\frac{1}{4} \ln(x^2 + 1) + \frac{1}{2} \tan^{-1}(x) + \frac{1}{2} \ln|x - 1| + C$

Visual Representation of Function Growth

Final Solution: $\frac{1}{2} \ln|x - 1| - \frac{1}{4} \ln(x^2 + 1) + \frac{1}{2} \tan^{-1}(x) + C$

Step 1: Decomposition into Partial Fractions

To integrate the rational function $I = \int \frac{3x-1}{(x+2)^2} \, dx$, we express the integrand as a sum of partial fractions. For a repeated linear factor in the denominator, the decomposition is given by:

$\frac{3x-1}{(x+2)^2} = \frac{A}{x+2} + \frac{B}{(x+2)^2}$ [By Partial Fraction Decomposition Theorem for repeated roots]

Step 2: Determining Constants A and B

Multiply both sides by $(x+2)^2$ to clear the denominators:

$3x - 1 = A(x + 2) + B$

$3x - 1 = Ax + 2A + B$

By comparing the coefficients of like powers of $x$ on both sides:

• Coefficient of $x$: $A = 3$
• Constant term: $2A + B = -1$

Substitute $A = 3$ into the constant term equation:

$2(3) + B = -1 \implies 6 + B = -1 \implies B = -7$

Step 3: Setting up the Integral

Substitute the constants $A$ and $B$ back into the integral expression:

$I = \int \left( \frac{3}{x+2} - \frac{7}{(x+2)^2} \right) \, dx$

$I = 3 \int \frac{1}{x+2} \, dx - 7 \int (x+2)^{-2} \, dx$

Step 4: Execution of Integration

Applying the standard integration rules:

• $\int \frac{1}{x} \, dx = \ln|x| + C$ [Per the logarithmic integration rule]
• $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$ [Per the Power Rule for Integration]

$I = 3 \ln|x+2| - 7 \left( \frac{(x+2)^{-2+1}}{-2+1} \right) + C$

$I = 3 \ln|x+2| - 7 \left( \frac{(x+2)^{-1}}{-1} \right) + C$

$I = 3 \ln|x+2| + \frac{7}{x+2} + C$

Final Solution: $3 \ln|x+2| + \frac{7}{x+2} + C$

Step 1: Decomposition via Partial Fractions

To integrate the rational function $f(x) = \frac{2}{(1-x)(1+x^2)}$, we express the integrand as a sum of partial fractions. Since $(1+x^2)$ is an irreducible quadratic factor, the partial fraction decomposition takes the form:

$\frac{2}{(1-x)(1+x^2)} = \frac{A}{1-x} + \frac{Bx + C}{1+x^2}$ [By the Theory of Partial Fraction Decomposition for distinct linear and irreducible quadratic factors]

Step 2: Determining the Constants

Equating the numerators after finding a common denominator:

$2 = A(1+x^2) + (Bx + C)(1-x)$

$2 = A + Ax^2 + Bx - Bx^2 + C - Cx$

$2 = (A - B)x^2 + (B - C)x + (A + C)$

By comparing coefficients of powers of $x$ on both sides:

• For $x^2$: $A - B = 0 \implies A = B$
• For $x^1$: $B - C = 0 \implies B = C$
• For $x^0$: $A + C = 2$

Substituting $A=B$ and $B=C$ into the constant term equation:

$A + A = 2 \implies 2A = 2 \implies A = 1$

Therefore, $A = 1, B = 1, C = 1$.

Step 3: Setting up the Integral

The integral becomes:

$\int \frac{2}{(1-x)(1+x^2)} dx = \int \frac{1}{1-x} dx + \int \frac{x+1}{1+x^2} dx$

Splitting the second integral:

$I = \int \frac{1}{1-x} dx + \int \frac{x}{1+x^2} dx + \int \frac{1}{1+x^2} dx$

Step 4: Evaluating the Components

• $\int \frac{1}{1-x} dx = -\ln|1-x| + C_1$ [Using substitution $u = 1-x, du = -dx$]
• $\int \frac{x}{1+x^2} dx = \frac{1}{2} \ln(1+x^2) + C_2$ [Using substitution $v = 1+x^2, dv = 2x dx$]
• $\int \frac{1}{1+x^2} dx = \tan^{-1}(x) + C_3$ [Per the standard integral formula for $\int \frac{1}{1+x^2}dx$]

Visual Representation of the Integrand

Final Synthesis

Combining the results and adding the constant of integration $C$:

$I = -\ln|1-x| + \frac{1}{2} \ln(1+x^2) + \tan^{-1}(x) + C$

Final Solution: -\ln|1-x| + \frac{1}{2} \ln(1+x^2) + \tan^{-1}(x) + C

Step 1: Analyzing the Degree of the Rational Function

Given the integrand $f(x) = \frac{x^3 + x + 1}{x^2 - 1}$. Observe that the degree of the numerator $P(x) = x^3 + x + 1$ is 3, and the degree of the denominator $Q(x) = x^2 - 1$ is 2. Since the degree of the numerator is greater than or equal to the degree of the denominator, we must perform polynomial long division to express the function in a simpler form [Per the Division Algorithm for Polynomials].

Step 2: Polynomial Long Division

Divide $x^3 + x + 1$ by $x^2 - 1$:

Thus, the expression can be rewritten as:

$\frac{x^3 + x + 1}{x^2 - 1} = x + \frac{2x + 1}{x^2 - 1}$

Step 3: Partial Fraction Decomposition

We decompose the proper fraction $\frac{2x + 1}{x^2 - 1}$ using partial fractions. Note that $x^2 - 1 = (x - 1)(x + 1)$.

$\frac{2x + 1}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$

Multiplying by the common denominator $(x - 1)(x + 1)$:

$2x + 1 = A(x + 1) + B(x - 1)$

• Set $x = 1$: $2(1) + 1 = A(1 + 1) \implies 3 = 2A \implies A = \frac{3}{2}$
• Set $x = -1$: $2(-1) + 1 = B(-1 - 1) \implies -1 = -2B \implies B = \frac{1}{2}$

Step 4: Integration of the Components

Substitute the partial fractions back into the integral:

$\int \frac{x^3 + x + 1}{x^2 - 1} dx = \int \left( x + \frac{3/2}{x - 1} + \frac{1/2}{x + 1} \right) dx$

Apply the linearity property of integrals [Per the Sum Rule of Integration]:

$= \int x dx + \frac{3}{2} \int \frac{1}{x - 1} dx + \frac{1}{2} \int \frac{1}{x + 1} dx$

Perform the integration:

$= \frac{x^2}{2} + \frac{3}{2} \ln|x - 1| + \frac{1}{2} \ln|x + 1| + C$

Visual Representation of the integrand behavior near asymptotes

Final Solution: $\frac{x^2}{2} + \frac{3}{2} \ln|x - 1| + \frac{1}{2} \ln|x + 1| + C$

Step 1: Substitution and Differential Transformation

To integrate $I = \int \frac{1}{e^x - 1} \, dx$, we apply the suggested substitution:

Let $e^x = t$ [Substitution Method].

Differentiating with respect to $x$: $\frac{d}{dx}(e^x) = \frac{dt}{dx} \implies e^x \, dx = dt$.

Since $e^x = t$, we have $dx = \frac{dt}{t}$.

Substituting these into the integral:

$I = \int \frac{1}{(t-1) \cdot t} \, dt$

Step 2: Partial Fraction Decomposition

We decompose the integrand $\frac{1}{t(t-1)}$ into simpler partial fractions [Method of Partial Fractions].

$\frac{1}{t(t-1)} = \frac{A}{t} + \frac{B}{t-1}$

Multiplying both sides by $t(t-1)$, we obtain the identity:

$1 = A(t-1) + Bt$

To find the constants $A$ and $B$:

• Setting $t = 0$: $1 = A(0-1) \implies A = -1$.
• Setting $t = 1$: $1 = B(1) \implies B = 1$.

Thus, the expression becomes:

$\frac{1}{t(t-1)} = -\frac{1}{t} + \frac{1}{t-1}$

Step 3: Integration and Back-Substitution

Integrate with respect to $t$:

$I = \int \left( -\frac{1}{t} + \frac{1}{t-1} \right) \, dt$

[By the linearity property of integrals]:

$I = -\int \frac{1}{t} \, dt + \int \frac{1}{t-1} \, dt$

$I = -\ln|t| + \ln|t-1| + C$ [Integral formula $\int \frac{1}{x} \, dx = \ln|x| + C$].

Using logarithm properties ($\ln a - \ln b = \ln \frac{a}{b}$):

$I = \ln \left| \frac{t-1}{t} \right| + C$

Substituting $t = e^x$ back into the expression:

$I = \ln \left| \frac{e^x - 1}{e^x} \right| + C$

This can be simplified as:

$I = \ln |1 - e^{-x}| + C$

Visual Representation of the Integrand

Final Solution: \ln|e^x - 1| - x + C \text{ or } \ln|1 - e^{-x}| + C

Step 1: Decomposition of the Rational Function

Given the integrand $\frac{3x-1}{(x-1)(x-2)(x-3)}$, we represent it as a sum of partial fractions [Per the Method of Partial Fractions for distinct linear factors]:

$$\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$$

Step 2: Solving for Coefficients A, B, and C

Equating the numerators:

$$3x - 1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$$

• To find $A$, set $x = 1$: $3(1)-1 = A(1-2)(1-3) \Rightarrow 2 = A(-1)(-2) \Rightarrow 2 = 2A \Rightarrow A = 1$
• To find $B$, set $x = 2$: $3(2)-1 = B(2-1)(2-3) \Rightarrow 5 = B(1)(-1) \Rightarrow 5 = -B \Rightarrow B = -5$
• To find $C$, set $x = 3$: $3(3)-1 = C(3-1)(3-2) \Rightarrow 8 = C(2)(1) \Rightarrow 8 = 2C \Rightarrow C = 4$

Step 3: Setting up the Integral

Substitute the constants back into the integral expression:

$$\int \frac{3x-1}{(x-1)(x-2)(x-3)} dx = \int \left( \frac{1}{x-1} - \frac{5}{x-2} + \frac{4}{x-3} \right) dx$$

Step 4: Execution of Integration

Apply the linearity property of integrals and the standard integral formula $\int \frac{1}{u} du = \ln|u| + C$ [Per Fundamental Theorem of Calculus]:

$$\int \frac{1}{x-1} dx - 5 \int \frac{1}{x-2} dx + 4 \int \frac{1}{x-3} dx$$

$$= \ln|x-1| - 5\ln|x-2| + 4\ln|x-3| + C$$

Step 5: Logarithmic Simplification

Utilizing the property $n \ln(m) = \ln(m^n)$ and $\ln(a) - \ln(b) + \ln(c) = \ln\left(\frac{ac}{b}\right)$:

$$= \ln|x-1| + \ln|(x-3)^4| - \ln|(x-2)^5| + C$$

$$= \ln\left| \frac{(x-1)(x-3)^4}{(x-2)^5} \right| + C$$

Final Solution: \ln|x-1| - 5\ln|x-2| + 4\ln|x-3| + C