Step 1: Formal Representation of the Integral

To evaluate the integral of the function $I = \int \frac{1}{\sqrt{9-25x^2}} \, dx$, we first identify the standard form required for integration involving quadratic radicals.

Step 2: Algebraic Manipulation and Factorization

We aim to bring the denominator into the standard form $\sqrt{a^2 - u^2}$. To achieve this, we factor out $25$ from the expression inside the square root:

$9 - 25x^2 = 25\left(\frac{9}{25} - x^2\right)$

Substituting this back into the integral:

$I = \int \frac{1}{\sqrt{25\left(\left(\frac{3}{5}\right)^2 - x^2\right)}} \, dx$

[By the property of radicals $\sqrt{A \cdot B} = \sqrt{A} \cdot \sqrt{B}$]

$I = \int \frac{1}{5\sqrt{\left(\frac{3}{5}\right)^2 - x^2}} \, dx$

$I = \frac{1}{5} \int \frac{1}{\sqrt{\left(\frac{3}{5}\right)^2 - x^2}} \, dx$

Step 3: Application of the Standard Integration Formula

We utilize the standard integration formula for the form $\int \frac{1}{\sqrt{a^2 - x^2}} \, dx$:

$\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C$

[Per the standard integral table for inverse trigonometric functions]

Here, $a = \frac{3}{5}$. Substituting these values into the expression:

$I = \frac{1}{5} \left( \arcsin\left(\frac{x}{3/5}\right) \right) + C$

Step 4: Simplification of the Result

Simplify the fraction within the inverse sine function:

$I = \frac{1}{5} \arcsin\left(\frac{5x}{3}\right) + C$

Visual Representation of the Integration Geometry

Final Solution: $\frac{1}{5} \arcsin\left(\frac{5x}{3}\right) + C$

Given Variables & Initial Setup

We are required to evaluate the indefinite integral:

$$I = \int \frac{x+3}{x^2-2x-5} \, dx$$

Step 1: Decomposition of the Numerator

To integrate this rational function, we express the numerator $(x+3)$ in terms of the derivative of the denominator $(x^2-2x-5)$. Let:

$$x+3 = A \cdot \frac{d}{dx}(x^2-2x-5) + B$$

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

Equating coefficients of $x$ and the constant terms [By the Principle of Undetermined Coefficients]:

• For $x$: $2A = 1 \implies A = \frac{1}{2}$
• For constants: $-2A + B = 3 \implies -2(\frac{1}{2}) + B = 3 \implies -1 + B = 3 \implies B = 4$

Thus, $x+3 = \frac{1}{2}(2x-2) + 4$.

Step 2: Splitting the Integral

Substitute the decomposition back into the original integral $I$:

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

$$I = \frac{1}{2} \int \frac{2x-2}{x^2-2x-5} \, dx + 4 \int \frac{1}{x^2-2x-5} \, dx$$

Let $I = \frac{1}{2}I_1 + 4I_2$.

Step 3: Evaluating $I_1$ and $I_2$

For $I_1 = \int \frac{2x-2}{x^2-2x-5} \, dx$, let $u = x^2-2x-5$, then $du = (2x-2)dx$ [By the Method of Substitution]:

$$I_1 = \int \frac{1}{u} \, du = \ln|u| + C_1 = \ln|x^2-2x-5| + C_1$$

For $I_2 = \int \frac{1}{x^2-2x-5} \, dx$, complete the square in the denominator:

$$x^2-2x-5 = (x^2-2x+1) - 1 - 5 = (x-1)^2 - (\sqrt{6})^2$$

Using the standard integral formula $\int \frac{1}{x^2-a^2} \, dx = \frac{1}{2a} \ln\left| \frac{x-a}{x+a} \right| + C$:

$$I_2 = \int \frac{1}{(x-1)^2 - (\sqrt{6})^2} \, dx = \frac{1}{2\sqrt{6}} \ln\left| \frac{(x-1)-\sqrt{6}}{(x-1)+\sqrt{6}} \right| + C_2$$

Step 4: Final Combination

Substitute $I_1$ and $I_2$ back into $I = \frac{1}{2}I_1 + 4I_2$:

$$I = \frac{1}{2} \ln|x^2-2x-5| + 4 \left( \frac{1}{2\sqrt{6}} \ln\left| \frac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right| \right) + C$$

Simplifying $4 \cdot \frac{1}{2\sqrt{6}} = \frac{2}{\sqrt{6}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3}$:

Final Solution: $\frac{1}{2} \ln|x^2-2x-5| + \frac{2}{\sqrt{6}} \ln\left| \frac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right| + C$

Step 1: Reformulating the Integral

The given integral is $I = \int \frac{x^2}{1 - x^6} \, dx$. To simplify this expression into a standard form, observe that $x^6 = (x^3)^2$. Thus, the integral becomes:

Step 2: Method of Substitution

Let $u = x^3$. Differentiating both sides with respect to $x$ yields:

[By the Method of Substitution for Integration]

Step 3: Transforming the Integral

Substitute $u$ and $du$ into the original expression:

Step 4: Applying Standard Integration Formula

Utilize the standard integral form: $\int \frac{1}{a^2 - x^2} \, dx = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C$. Here, $a = 1$ and $x = u$.

Step 5: Back-Substitution

Substitute $u = x^3$ back into the result to obtain the final expression in terms of $x$:

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

Step 1: Decomposition of the Integrand

We are required to evaluate the integral: $I = \int \frac{x+2}{\sqrt{x^2+2x+3}} \, dx$

To solve this, we express the numerator $(x+2)$ in terms of the derivative of the quadratic expression under the square root, $f(x) = x^2+2x+3$. The derivative is $f'(x) = 2x+2$.

We set: $x+2 = A \frac{d}{dx}(x^2+2x+3) + B$

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

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

Comparing coefficients of $x$ and constants [By Principle of Equating Coefficients]:

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

Step 2: Splitting the Integral

Rewrite the original integral using these constants:

Let $I = \frac{1}{2} I_1 + I_2$, where $I_1 = \int \frac{2x+2}{\sqrt{x^2+2x+3}} \, dx$ and $I_2 = \int \frac{1}{\sqrt{x^2+2x+3}} \, dx$.

Step 3: Solving $I_1$

For $I_1$, let $u = x^2+2x+3$, then $du = (2x+2)dx$ [By Method of Substitution].

Step 4: Solving $I_2$

For $I_2$, complete the square for the denominator: $x^2+2x+3 = (x^2+2x+1) + 2 = (x+1)^2 + (\sqrt{2})^2$.

Using the standard integration formula $\int \frac{dx}{\sqrt{x^2+a^2}} = \ln|x + \sqrt{x^2+a^2}| + C$:

Step 5: Final Compilation

$I = \frac{1}{2} (2\sqrt{x^2+2x+3}) + \ln|x+1 + \sqrt{x^2+2x+3}| + C$

$I = \sqrt{x^2+2x+3} + \ln|x+1 + \sqrt{x^2+2x+3}| + C$

Final Solution: \sqrt{x^2+2x+3} + \ln|x+1 + \sqrt{x^2+2x+3}| + C

Step 1: Analyzing the Integrand

The objective is to evaluate the integral:

$I = \int \frac{3x}{1 + 2x^4} \, dx$

Observe that the numerator contains $x$, which suggests a substitution involving the derivative of $x^2$. We rewrite the denominator to express $x^4$ as a square: $2x^4 = (\sqrt{2}x^2)^2$.

Step 2: Substitution Method

Let $u = \sqrt{2}x^2$. [By the Method of Substitution]

Differentiating with respect to $x$:

$\frac{du}{dx} = 2\sqrt{2}x \implies dx = \frac{du}{2\sqrt{2}x}$

Step 3: Transforming the Integral

Substitute $u$ and $dx$ into the original integral:

$I = \int \frac{3x}{1 + u^2} \cdot \frac{du}{2\sqrt{2}x}$

Canceling $x$ from the numerator and denominator:

$I = \frac{3}{2\sqrt{2}} \int \frac{1}{1 + u^2} \, du$

Step 4: Integration Using Standard Formula

Apply the standard integral identity: $\int \frac{1}{1+t^2} \, dt = \tan^{-1}(t) + C$. [Per the Fundamental Theorem of Calculus]

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

Step 5: Back-Substitution

Substitute $u = \sqrt{2}x^2$ back into the expression:

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

Step 6: Rationalizing the Coefficient (Optional but Standard)

Multiply the numerator and denominator of the coefficient by $\sqrt{2}$:

$\frac{3}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{4}$

$I = \frac{3\sqrt{2}}{4} \tan^{-1}(\sqrt{2}x^2) + C$

Visual Representation of the Integrand Function

Final Solution: $\frac{3}{2\sqrt{2}} \tan^{-1}(\sqrt{2}x^2) + C$ or $\frac{3\sqrt{2}}{4} \tan^{-1}(\sqrt{2}x^2) + C$

Given Integral

Evaluate the integral: $I = \int \frac{4x+1}{\sqrt{2x^2+x-3}} \, dx$

Step 1: Expressing the Numerator as a Derivative of the Quadratic Under the Radical

Let $f(x) = 2x^2 + x - 3$. The derivative of $f(x)$ is $f'(x) = 4x + 1$. [By the substitution rule for differentiation].

We observe that the numerator is exactly the derivative of the expression inside the square root. Therefore, we can express the integral in terms of $f(x)$ and $f'(x)$.

Step 2: Substitution Method

Let $t = 2x^2 + x - 3$.

Differentiating both sides with respect to $x$: $\frac{dt}{dx} = 4x + 1 \implies dt = (4x + 1) \, dx$. [By the Chain Rule of Calculus].

Substituting these into the original integral:

$I = \int \frac{dt}{\sqrt{t}}$

$I = \int t^{-1/2} \, dt$

Step 3: Evaluating the Integral

Apply the power rule for integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$. [By the Fundamental Theorem of Calculus].

$I = \frac{t^{-1/2 + 1}}{-1/2 + 1} + C$

$I = \frac{t^{1/2}}{1/2} + C$

$I = 2\sqrt{t} + C$

Step 4: Back-Substitution

Substitute $t = 2x^2 + x - 3$ back into the expression:

$I = 2\sqrt{2x^2 + x - 3} + C$

Visual Representation of the Function

Final Solution: $2\sqrt{2x^2+x-3} + C$

Step 1: Decomposition of the Integrand

Given the integral: $I = \int \frac{5x+3}{\sqrt{x^2+4x+10}} \, dx$.
To solve this, we express the numerator $(5x+3)$ in terms of the derivative of the quadratic expression inside the square root, $f(x) = x^2+4x+10$.
The derivative $f'(x) = 2x+4$.
We set up the linear combination: $5x+3 = A(2x+4) + B$.

Step 2: Determining Constants A and B

Equating the coefficients of $x$ and the constant terms:
$5x + 3 = 2Ax + 4A + B$
Comparing $x$ coefficients: $2A = 5 \implies A = \frac{5}{2}$.
Comparing constants: $4A + B = 3 \implies 4\left(\frac{5}{2}\right) + B = 3 \implies 10 + B = 3 \implies B = -7$.
Thus, the integrand becomes: $\frac{5}{2} \cdot \frac{2x+4}{\sqrt{x^2+4x+10}} - \frac{7}{\sqrt{x^2+4x+10}}$.

Step 3: Evaluating the Two Integrals

The integral splits into $I = \frac{5}{2} I_1 - 7 I_2$, where:
$I_1 = \int \frac{2x+4}{\sqrt{x^2+4x+10}} \, dx$ and $I_2 = \int \frac{1}{\sqrt{x^2+4x+10}} \, dx$.

For $I_1$, let $u = x^2+4x+10$, then $du = (2x+4)dx$ [By Method of Substitution].
$I_1 = \int u^{-1/2} \, du = 2u^{1/2} = 2\sqrt{x^2+4x+10}$.

Step 4: Completing the Square for $I_2$

Complete the square for the denominator: $x^2+4x+10 = (x^2+4x+4) + 6 = (x+2)^2 + (\sqrt{6})^2$.
Using the standard integral formula $\int \frac{1}{\sqrt{x^2+a^2}} dx = \ln|x + \sqrt{x^2+a^2}| + C$ [By Integral of Algebraic Functions]:
$I_2 = \int \frac{dx}{\sqrt{(x+2)^2 + (\sqrt{6})^2}} = \ln|(x+2) + \sqrt{(x+2)^2+6}|$.
$I_2 = \ln|x+2 + \sqrt{x^2+4x+10}|$.

Step 5: Final Synthesis

Combining the results:
$I = \frac{5}{2} \left( 2\sqrt{x^2+4x+10} \right) - 7 \left( \ln|x+2 + \sqrt{x^2+4x+10}| \right) + C$.

Final Solution: $5\sqrt{x^2+4x+10} - 7\ln|x+2 + \sqrt{x^2+4x+10}| + C$

Step 1: Decomposition of the Integrand

The given integral is $I = \int \frac{x-1}{\sqrt{x^2-1}} dx$. We split the integral into two distinct parts to facilitate different integration techniques [By the linearity property of integration]:

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

Let $I_1 = \int \frac{x}{\sqrt{x^2-1}} dx$ and $I_2 = \int \frac{1}{\sqrt{x^2-1}} dx$. Thus, $I = I_1 - I_2$.

Step 2: Solving $I_1$ via Substitution

For $I_1 = \int \frac{x}{\sqrt{x^2-1}} dx$, we utilize the method of substitution [By the substitution rule for indefinite integrals]:

• Let $u = x^2 - 1$.
• Differentiating with respect to $x$, $du = 2x \, dx$, which implies $x \, dx = \frac{du}{2}$.

$I_1 = \int \frac{1}{\sqrt{u}} \cdot \frac{du}{2} = \frac{1}{2} \int u^{-1/2} du$

Applying the power rule $\int u^n du = \frac{u^{n+1}}{n+1} + C$:

$I_1 = \frac{1}{2} \left( \frac{u^{1/2}}{1/2} \right) = \sqrt{u} + C_1 = \sqrt{x^2-1} + C_1$

Step 3: Solving $I_2$ via Standard Formula

For $I_2 = \int \frac{1}{\sqrt{x^2-1}} dx$, we apply the standard integral formula [Per the table of integrals for algebraic forms]:

$\int \frac{1}{\sqrt{x^2-a^2}} dx = \ln|x + \sqrt{x^2-a^2}| + C$

Given $a = 1$, we have:

$I_2 = \ln|x + \sqrt{x^2-1}| + C_2$

Step 4: Synthesis of the Final Result

Combining the results from Step 2 and Step 3 into our expression from Step 1:

$I = I_1 - I_2 = \sqrt{x^2-1} - \ln|x + \sqrt{x^2-1}| + C$

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

Given Integral

We are required to evaluate the following indefinite integral: $I = \int \frac{1}{\sqrt{7 - 6x - x^2}} \, dx$

Step 1: Completing the Square in the Quadratic Expression

To integrate functions of the form $\frac{1}{\sqrt{ax^2 + bx + c}}$, we must express the quadratic $7 - 6x - x^2$ in the form $a^2 - (x+h)^2$ [By the method of completing the square].

Consider the expression: $-(x^2 + 6x - 7)$

To complete the square for $x^2 + 6x$, we add and subtract the square of half the coefficient of $x$, which is $(\frac{6}{2})^2 = 3^2 = 9$.

$-(x^2 + 6x + 9 - 9 - 7)$

$-( (x+3)^2 - 16 )$

$16 - (x+3)^2 = 4^2 - (x+3)^2$

Step 2: Substituting into the Integral

Substituting the completed square back into the integral:

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

Step 3: Applying the Standard Integral Formula

We utilize the standard integration formula [Per the Fundamental Theorem of Calculus/Integration]:

$\int \frac{1}{\sqrt{a^2 - u^2}} \, du = \sin^{-1}\left(\frac{u}{a}\right) + C$

Here, $a = 4$ and $u = (x+3)$. Consequently, $du = dx$.

$I = \sin^{-1}\left(\frac{x+3}{4}\right) + C$

Visual Representation of the Quadratic Geometry

Step 4: Final Verification

To verify the result, differentiate $I = \sin^{-1}\left(\frac{x+3}{4}\right)$ with respect to $x$ using the chain rule:

$\frac{d}{dx} [\sin^{-1}(\frac{x+3}{4})] = \frac{1}{\sqrt{1 - (\frac{x+3}{4})^2}} \cdot \frac{1}{4}$

$= \frac{1}{\sqrt{\frac{16 - (x^2 + 6x + 9)}{16}}} \cdot \frac{1}{4}$

$= \frac{1}{\frac{\sqrt{7 - 6x - x^2}}{4}} \cdot \frac{1}{4} = \frac{1}{\sqrt{7 - 6x - x^2}}$

The derivative matches the integrand, confirming the solution is correct.

Final Solution: $\sin^{-1}\left(\frac{x+3}{4}\right) + C$

Step 1: Identifying the Form and Substitution

The given integral is: $I = \int \frac{3x^2}{x^6 + 1} \, dx$

Observe that the integrand contains $x^6$ in the denominator, which can be expressed as $(x^3)^2$. We note that the derivative of $x^3$ is $3x^2$, which is present in the numerator. This suggests the method of integration by substitution [Per the Substitution Rule for Integrals].

Step 2: Performing the Variable Transformation

Let $u = x^3$.

Differentiating with respect to $x$: $\frac{du}{dx} = 3x^2$ $du = 3x^2 \, dx$ [By the Chain Rule of differentiation]

Step 3: Rewriting the Integral

Substituting $u$ and $du$ into the original integral:

Step 4: Applying the Standard Integral Formula

The integral of the form $\int \frac{1}{x^2 + a^2} \, dx$ is given by the standard formula: $\int \frac{1}{u^2 + 1^2} \, du = \tan^{-1}(u) + C$ [Where $C$ is the constant of integration]

Step 5: Final Substitution and Result

Substitute $u = x^3$ back into the result to return to the original variable $x$:

Visual Representation of the Integrand Mapping

Final Solution: $\tan^{-1}(x^3) + C$

Step 1: Identifying the Form

The integral to be evaluated is $I = \int \frac{x+2}{\sqrt{4x-x^2}} dx$. This follows the standard form $\int \frac{px+q}{\sqrt{ax^2+bx+c}} dx$. To solve this, we express the numerator in terms of the derivative of the quadratic expression under the square root [Algebraic manipulation technique for integration].

Step 2: Decomposition of the Numerator

Let the derivative of the quadratic expression $4x - x^2$ be $f'(x) = 4 - 2x$. We set the numerator $x+2 = A \cdot \frac{d}{dx}(4x-x^2) + B$.

$x + 2 = A(4 - 2x) + B$

$x + 2 = -2Ax + (4A + B)$

By equating the coefficients of $x$ and the constant terms on both sides:

• $-2A = 1 \implies A = -\frac{1}{2}$
• $4A + B = 2 \implies 4(-\frac{1}{2}) + B = 2 \implies -2 + B = 2 \implies B = 4$

Thus, $x+2 = -\frac{1}{2}(4-2x) + 4$.

Step 3: Rewriting the Integral

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

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

Let $I = -\frac{1}{2} I_1 + 4 I_2$.

Step 4: Evaluating $I_1$

For $I_1 = \int \frac{4-2x}{\sqrt{4x-x^2}} dx$, let $u = 4x - x^2$. Then $du = (4-2x) dx$.

$I_1 = \int u^{-1/2} du = \frac{u^{1/2}}{1/2} = 2\sqrt{4x-x^2}$ [By Power Rule of Integration].

Step 5: Evaluating $I_2$

For $I_2 = \int \frac{1}{\sqrt{4x-x^2}} dx$, complete the square in the denominator:

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

$I_2 = \int \frac{1}{\sqrt{2^2 - (x-2)^2}} dx$.

Using the standard formula $\int \frac{1}{\sqrt{a^2-x^2}} dx = \sin^{-1}(\frac{x}{a}) + C$, we obtain:

$I_2 = \sin^{-1}\left(\frac{x-2}{2}\right)$.

Step 6: Final Combination

$I = -\frac{1}{2}(2\sqrt{4x-x^2}) + 4(\sin^{-1}(\frac{x-2}{2})) + C$

$I = -\sqrt{4x-x^2} + 4\sin^{-1}\left(\frac{x-2}{2}\right) + C$.

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

Step 1: Identifying the Form of the Integral

The given integral is of the form:

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

This matches the standard integration formula: $\int \frac{1}{\sqrt{x^2 + a^2}} \, dx = \ln|x + \sqrt{x^2 + a^2}| + C$ [Per the standard integral for inverse hyperbolic functions/logarithmic forms].

Step 2: Substitution and Simplification

Let $u = 2 - x$. Then, differentiating both sides with respect to $x$:

$\frac{du}{dx} = -1 \implies dx = -du$

Substituting these values into the integral:

$I = \int \frac{1}{\sqrt{u^2 + 1}} \cdot (-du)$

$I = -\int \frac{1}{\sqrt{u^2 + 1}} \, du$

Step 3: Applying the Standard Formula

Using the formula $\int \frac{1}{\sqrt{u^2 + a^2}} \, du = \ln|u + \sqrt{u^2 + a^2}| + C$, where $a = 1$:

$I = - \ln|u + \sqrt{u^2 + 1}| + C$

Step 4: Back-Substitution

Substitute $u = 2 - x$ back into the expression:

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

Step 5: Refinement of the Result

Using the logarithmic property $-\ln|A| = \ln|\frac{1}{A}|$, we can rewrite the expression, or keep it in the current standard form. Expanding the square inside the radical for clarity:

$(2-x)^2 + 1 = 4 - 4x + x^2 + 1 = x^2 - 4x + 5$

Thus, $I = -\ln|2 - x + \sqrt{x^2 - 4x + 5}| + C$

Final Solution: $-\ln|2 - x + \sqrt{x^2 - 4x + 5}| + C$

Step 1: Expanding the Quadratic Expression

To integrate the function $I = \int \frac{1}{\sqrt{(x-1)(x-2)}} \, dx$, we first simplify the quadratic expression under the square root:

$(x-1)(x-2) = x^2 - 2x - x + 2 = x^2 - 3x + 2$

[By algebraic expansion of binomials]

Step 2: Completing the Square

To express the quadratic in the form $(x-a)^2 - b^2$, we perform the method of completing the square:

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

$= (x - \frac{3}{2})^2 - \frac{9}{4} + 2$

$= (x - \frac{3}{2})^2 - \frac{1}{4}$

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

[By algebraic identity $(a-b)^2 = a^2 - 2ab + b^2$]

Step 3: Applying the Standard Integral Formula

Substitute the completed square back into the integral:

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

We use the standard integral formula: $\int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \ln |x + \sqrt{x^2 - a^2}| + C$

[Per standard integration table for algebraic radicals]

Step 4: Final Evaluation

Applying the formula where $x = (x - \frac{3}{2})$ and $a = \frac{1}{2}$:

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

Since $\sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2} = \sqrt{x^2 - 3x + 2}$, we substitute back:

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

This can also be written as:

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

Using the property $\ln(\frac{A}{B}) = \ln A - \ln B$, the constant term $-\ln(2)$ can be absorbed into the arbitrary constant $C$:

$I = \ln |2x - 3 + 2\sqrt{x^2 - 3x + 2}| + C_1$

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

Step 1: Identifying the Form and Substitution

To evaluate the integral $I = \int \frac{1}{\sqrt{1 + 4x^2}} \, dx$, we observe that the denominator is of the form $\sqrt{a^2 + u^2}$. First, we rewrite the expression to isolate the variable $x$ in a squared form:

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

We apply the method of substitution: Let $u = 2x$. [By Substitution Rule].

Differentiating with respect to $x$: $\frac{du}{dx} = 2 \implies dx = \frac{du}{2}$.

Step 2: Transforming the Integral

Substitute $u$ and $dx$ into the original integral expression:

$I = \int \frac{1}{\sqrt{1 + u^2}} \cdot \frac{du}{2}$

$I = \frac{1}{2} \int \frac{1}{\sqrt{a^2 + u^2}} \, du$ where $a = 1$.

Step 3: Applying Standard Integration Formula

We utilize the standard integration formula for the reciprocal of the square root of a quadratic sum: $\int \frac{1}{\sqrt{x^2 + a^2}} \, dx = \ln|x + \sqrt{x^2 + a^2}| + C$. [Per Standard Integral Identity].

Applying this to our transformed integral:

$I = \frac{1}{2} \ln|u + \sqrt{u^2 + 1}| + C$

Step 4: Re-substituting and Simplifying

Substitute $u = 2x$ back into the result to return to the original variable:

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

$I = \frac{1}{2} \ln|2x + \sqrt{4x^2 + 1}| + C$

Final Solution: $\frac{1}{2} \ln|2x + \sqrt{4x^2 + 1}| + C$

Step 1: Analyzing the Integral Form

The objective is to compute the integral:

$I = \int \frac{1}{\sqrt{8+3x-x^2}} \, dx$

To integrate functions of the form $\frac{1}{\sqrt{ax^2+bx+c}}$, we must convert the quadratic expression $ax^2+bx+c$ into the form $-(x-h)^2 + k^2$ via the method of completing the square [Algebraic manipulation of quadratics].

Step 2: Completing the Square

The quadratic expression is $Q(x) = -x^2 + 3x + 8$.

First, factor out the negative sign from the variable terms:

$Q(x) = -(x^2 - 3x) + 8$

To complete the square for $x^2 - 3x$, we add and subtract the square of half the coefficient of $x$. The coefficient of $x$ is $-3$, so half of it is $-\frac{3}{2}$. Its square is $\frac{9}{4}$.

$Q(x) = -(x^2 - 3x + \frac{9}{4} - \frac{9}{4}) + 8$

$Q(x) = -(x - \frac{3}{2})^2 + \frac{9}{4} + 8$

Convert $8$ to a fraction with denominator $4$: $8 = \frac{32}{4}$.

$Q(x) = \frac{41}{4} - (x - \frac{3}{2})^2$

$Q(x) = (\frac{\sqrt{41}}{2})^2 - (x - \frac{3}{2})^2$

Step 3: Applying the Integration Formula

Substituting back into the integral:

$I = \int \frac{1}{\sqrt{(\frac{\sqrt{41}}{2})^2 - (x - \frac{3}{2})^2}} \, dx$

Utilizing the standard integration formula [Integral of Inverse Trigonometric Function]:

$\int \frac{1}{\sqrt{a^2 - u^2}} \, du = \sin^{-1}(\frac{u}{a}) + C$

Here, $a = \frac{\sqrt{41}}{2}$ and $u = (x - \frac{3}{2})$.

Step 4: Final Evaluation

$I = \sin^{-1}\left( \frac{x - 3/2}{\sqrt{41}/2} \right) + C$

Simplify the fraction inside the $\sin^{-1}$ function:

$I = \sin^{-1}\left( \frac{\frac{2x-3}{2}}{\frac{\sqrt{41}}{2}} \right) + C$

$I = \sin^{-1}\left( \frac{2x-3}{\sqrt{41}} \right) + C$

Visual Representation of the Integrand

Final Solution: $\sin^{-1}\left( \frac{2x-3}{\sqrt{41}} \right) + C$

Step 1: Expanding the Quadratic Form

To integrate $I = \int \frac{1}{\sqrt{(x-a)(x-b)}} \, dx$, we must first express the product under the square root as a quadratic polynomial in $x$.

Expand the product: $(x-a)(x-b) = x^2 - (a+b)x + ab$.

Thus, $I = \int \frac{1}{\sqrt{x^2 - (a+b)x + ab}} \, dx$. [By Algebraic Expansion]

Step 2: Completing the Square

To evaluate the integral, transform the quadratic expression into the form $(x+k)^2 + c$.

Take the coefficient of $x$, which is $-(a+b)$. Divide by 2 and square it: $\left(\frac{a+b}{2}\right)^2$.

Add and subtract this value inside the square root:

$x^2 - (a+b)x + \left(\frac{a+b}{2}\right)^2 - \left(\frac{a+b}{2}\right)^2 + ab$

$= \left(x - \frac{a+b}{2}\right)^2 - \left[\frac{a^2 + 2ab + b^2 - 4ab}{4}\right]$

$= \left(x - \frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2$

Step 3: Applying Standard Integral Formulas

The integral now takes the form $\int \frac{dx}{\sqrt{X^2 - A^2}}$, where $X = \left(x - \frac{a+b}{2}\right)$ and $A = \left(\frac{a-b}{2}\right)$.

We use the standard formula: $\int \frac{1}{\sqrt{X^2 - A^2}} \, dX = \ln|X + \sqrt{X^2 - A^2}| + C$. [Per the Standard Integral Table]

Step 4: Final Substitution

Substituting $X$ and $A$ back into the logarithmic form:

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

Simplifying the expression within the square root back to $(x-a)(x-b)$:

$I = \ln \left| \left(x - \frac{a+b}{2}\right) + \sqrt{(x-a)(x-b)} \right| + C$

Final Solution: $\ln \left| \left(x - \frac{a+b}{2}\right) + \sqrt{(x-a)(x-b)} \right| + C$

Step 1: Analyzing the Quadratic Denominator

The given integral is $I = \int \frac{dx}{x^2 + 2x + 2}$. To integrate this, we must complete the square for the quadratic expression $x^2 + 2x + 2$.

The general form for completing the square of $x^2 + bx$ is to add and subtract $(\frac{b}{2})^2$. Here, $b = 2$, so we add and subtract $(\frac{2}{2})^2 = 1^2 = 1$.

$x^2 + 2x + 2 = (x^2 + 2x + 1) + 1$ [Algebraic completion of square]

$x^2 + 2x + 2 = (x + 1)^2 + 1$

Step 2: Substitution and Standard Integral Form

Rewrite the integral using the completed square:

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

Let $u = x + 1$. Then the differential $du = dx$ [Method of substitution].

The integral transforms to the standard form:

$I = \int \frac{du}{u^2 + a^2}$, where $a = 1$.

Step 3: Applying the Standard Integration Formula

We utilize the standard integration formula for inverse trigonometric functions:

$\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$ [Per the table of standard integrals]

Substituting $u$ and $a=1$ into the formula:

$I = \frac{1}{1} \tan^{-1}\left(\frac{u}{1}\right) + C$

$I = \tan^{-1}(u) + C$

Step 4: Back-Substitution

Replacing $u$ with the original variable expression $(x + 1)$:

$I = \tan^{-1}(x + 1) + C$

Final Solution: $\tan^{-1}(x + 1) + C$

Step 1: Analyzing the Integrand and Substitution Setup

The integral to be solved is:

$I = \int \frac{x^2}{\sqrt{x^6 + a^6}} \, dx$

To simplify the integral, we recognize that the numerator contains $x^2$, which is proportional to the derivative of $x^3$. We perform a substitution to transform the integral into a standard form. [By Substitution Rule for Integrals]

Let $u = x^3$.

Differentiating both sides with respect to $x$:

$\frac{du}{dx} = 3x^2 \implies du = 3x^2 \, dx \implies x^2 \, dx = \frac{du}{3}$

Step 2: Transforming the Integral

Substituting $u = x^3$ into the original integral, noting that $x^6 = (x^3)^2 = u^2$:

$I = \int \frac{1}{\sqrt{u^2 + (a^3)^2}} \cdot \frac{du}{3}$

$I = \frac{1}{3} \int \frac{du}{\sqrt{u^2 + (a^3)^2}}$

Step 3: Applying Standard Integral Formula

We utilize the standard integration formula for the form $\int \frac{dx}{\sqrt{x^2 + k^2}}$: [Standard Integral Identity]

$\int \frac{dx}{\sqrt{x^2 + k^2}} = \ln|x + \sqrt{x^2 + k^2}| + C$

In our expression, $x$ corresponds to $u$ and $k$ corresponds to $a^3$. Applying the formula:

$I = \frac{1}{3} \ln|u + \sqrt{u^2 + (a^3)^2}| + C$

Step 4: Back-Substitution

Substituting $u = x^3$ back into the result to return to the original variable $x$:

$I = \frac{1}{3} \ln|x^3 + \sqrt{(x^3)^2 + (a^3)^2}| + C$

$I = \frac{1}{3} \ln|x^3 + \sqrt{x^6 + a^6}| + C$

Visual Representation of the Transformation

Final Solution: \frac{1}{3} \ln|x^3 + \sqrt{x^6 + a^6}| + C

Step 1: Standardizing the Integrand

Given the integral: $I = \int \frac{dx}{\sqrt{9x - 4x^2}}$

To integrate this expression, we must complete the square for the quadratic expression under the radical: $f(x) = -4x^2 + 9x$.

First, factor out the coefficient of $x^2$, which is $-4$:

$f(x) = -4(x^2 - \frac{9}{4}x)$ [By Algebraic Factoring]

Now, complete the square by adding and subtracting $(\frac{1}{2} \cdot \frac{9}{4})^2 = (\frac{9}{8})^2 = \frac{81}{64}$ inside the parenthesis:

$f(x) = -4(x^2 - \frac{9}{4}x + \frac{81}{64} - \frac{81}{64})$

$f(x) = -4((x - \frac{9}{8})^2 - \frac{81}{64})$

$f(x) = 4(\frac{81}{64} - (x - \frac{9}{8})^2)$ [Distributing the $-4$]

$f(x) = (\frac{9}{2})^2 - 4(x - \frac{9}{8})^2$ is incorrect; rather, $f(x) = (\frac{9}{4})^2 - (2x - \frac{9}{4})^2$ is not the path. Let us retain the constant: $f(x) = (\frac{9}{4})^2 - (2x - \frac{9}{4})^2$. Actually, simplifying correctly: $f(x) = (\frac{9}{4})^2 - (2x - \frac{9}{4})^2$ is equivalent to $\frac{81}{16} - (4x^2 - 9x + \frac{81}{16}) = -4x^2 + 9x$.

Step 2: Substituting into the Integral

Substitute the completed square into the integral:

$I = \int \frac{dx}{\sqrt{(\frac{9}{4})^2 - (2x - \frac{9}{4})^2}}$

Let $u = 2x - \frac{9}{4}$, then $du = 2 dx \implies dx = \frac{du}{2}$.

$I = \frac{1}{2} \int \frac{du}{\sqrt{(\frac{9}{4})^2 - u^2}}$

Step 3: Applying the Standard Integral Formula

Utilize the standard integral formula: $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + C$ [Per the table of Standard Integrals].

$I = \frac{1}{2} \sin^{-1}(\frac{u}{9/4}) + C$

$I = \frac{1}{2} \sin^{-1}(\frac{4u}{9}) + C$

Step 4: Back-Substitution and Final Simplification

Substitute $u = 2x - \frac{9}{4} = \frac{8x - 9}{4}$ back into the expression:

$I = \frac{1}{2} \sin^{-1}(\frac{4(\frac{8x - 9}{4})}{9}) + C$

$I = \frac{1}{2} \sin^{-1}(\frac{8x - 9}{9}) + C$

Geometric Representation of the Integral Area

Final Solution: $\frac{1}{2} \sin^{-1}\left(\frac{8x-9}{9}\right) + C$

Step 1: Given Integral and Variable Substitution

We are required to evaluate the indefinite integral:

$I = \int \frac{\sec^2 x}{\sqrt{\tan^2 x + 4}} \, dx$

To simplify the integrand, we perform a substitution. Let $u = \tan x$. [Per the Method of Substitution for Integration]

Differentiating both sides with respect to $x$:

$\frac{du}{dx} = \sec^2 x \implies du = \sec^2 x \, dx$

Step 2: Transforming the Integral

Substituting $u$ and $du$ into the original integral, we obtain:

$I = \int \frac{1}{\sqrt{u^2 + 4}} \, du$

This expression takes the standard form $\int \frac{1}{\sqrt{u^2 + a^2}} \, du$, where $a^2 = 4$, implying $a = 2$.

Step 3: Applying the Standard Integration Formula

Using the standard integral formula [Reference: $\int \frac{1}{\sqrt{x^2 + a^2}} dx = \ln|x + \sqrt{x^2 + a^2}| + C$]:

$I = \ln|u + \sqrt{u^2 + 4}| + C$

Step 4: Back-Substitution

Reverting the variable $u$ back to the original trigonometric function $\tan x$:

$I = \ln|\tan x + \sqrt{\tan^2 x + 4}| + C$

Visual Representation of the Substitution Geometry

The transformation maps the trigonometric ratio to a standard algebraic form, allowing for the application of logarithmic integration properties derived from the inverse hyperbolic functions.

Final Solution: $\ln|\tan x + \sqrt{\tan^2 x + 4}| + C$