Step 1: Reformulation of the Integrand

Given the integral:

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

We simplify the integrand by distributing the division by $x$ across the term $(x+1)$:

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

Which reduces to:

$I = \int \left( 1 + \frac{1}{x} \right) (x + \log x)^2 \, dx$ [Algebraic simplification via the distributive property]

Step 2: Substitution Methodology

To evaluate the integral, we perform a change of variables. Let $u$ be the base of the squared term:

$u = x + \log x$

Now, compute the differential $du$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(\log x)$ $\frac{du}{dx} = 1 + \frac{1}{x}$ [By the Sum Rule of differentiation and the Derivative of $\log_e x$]

Rearranging gives:

$du = \left( 1 + \frac{1}{x} \right) \, dx$

Step 3: Evaluating the Transformed Integral

Substitute $u$ and $du$ into the integral expression derived in Step 1:

$I = \int (u)^2 \, du$

Applying the Power Rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, where $n=2$:

$I = \frac{u^{2+1}}{2+1} + C$ $I = \frac{u^3}{3} + C$ [By the Fundamental Theorem of Calculus]

Step 4: Reversion to Original Variable

Substitute $u = x + \log x$ back into the result to obtain the final expression in terms of $x$:

$I = \frac{(x + \log x)^3}{3} + C$

Final Solution: \frac{(x + \log x)^3}{3} + C

Step 1: Problem Definition and Algebraic Rearrangement

We are tasked to evaluate the following indefinite integral:

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

Utilizing the trigonometric identity $\frac{1}{\cos^2 x} = \sec^2 x$ [Per the Reciprocal Identity of Trigonometry], we rewrite the integrand to facilitate substitution:

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

Step 2: Method of Substitution

Let $u = 1 - \tan x$. To find the differential $du$, we differentiate $u$ with respect to $x$ [By the Chain Rule]:

$\frac{du}{dx} = \frac{d}{dx}(1 - \tan x) = 0 - \sec^2 x$

$du = -\sec^2 x \, dx$

Consequently, $\sec^2 x \, dx = -du$.

Step 3: Variable Transformation and Integration

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

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

Applying the Power Rule for Integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ [Where $n \neq -1$]:

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

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

$I = \frac{1}{u} + C$

Step 4: Back-Substitution

Reverting the variable $u$ to the original expression $1 - \tan x$:

$I = \frac{1}{1 - \tan x} + C$

Visual Representation of the Substitution Logic

Analytical Verification

Expression	Result
Integral term	$\int (1-\tan x)^{-2} \sec^2 x \, dx$
Resulting primitive	$(1-\tan x)^{-1} + C$
Derivative check	$-1(1-\tan x)^{-2} \cdot (-\sec^2 x) = \frac{\sec^2 x}{(1-\tan x)^2}$

Final Solution: $\frac{1}{1 - \tan x} + C$

Step 1: Setting up the Integral and Identifying the Substitution

We are tasked with evaluating the indefinite integral:

$I = \int \frac{x^3 \sin(\tan^{-1} x^4)}{1+x^8} \, dx$

To simplify the integrand, we observe the composition of functions. Let $u = \tan^{-1}(x^4)$. [Per the chain rule of differentiation, $\frac{d}{dx} \tan^{-1}(f(x)) = \frac{1}{1+(f(x))^2} \cdot f'(x)$]

Step 2: Performing the Differential Substitution

Let $u = \tan^{-1}(x^4)$.

Differentiating both sides with respect to $x$:

$\frac{du}{dx} = \frac{1}{1 + (x^4)^2} \cdot \frac{d}{dx}(x^4)$

$\frac{du}{dx} = \frac{1}{1 + x^8} \cdot 4x^3$

Rearranging the terms to solve for the differential $dx$:

$du = \frac{4x^3}{1+x^8} \, dx \implies \frac{x^3}{1+x^8} \, dx = \frac{1}{4} \, du$

Step 3: Transforming the Integral

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

$I = \int \sin(u) \cdot \frac{1}{4} \, du$

$I = \frac{1}{4} \int \sin(u) \, du$

Step 4: Integration and Back-Substitution

Using the fundamental integration formula $\int \sin(u) \, du = -\cos(u) + C$, where $C$ is the constant of integration:

$I = \frac{1}{4} (-\cos(u)) + C$

$I = -\frac{1}{4} \cos(u) + C$

Substituting the original variable $u = \tan^{-1}(x^4)$ back into the result:

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

Step 5: Geometric Interpretation of $\cos(\tan^{-1} x^4)$

Let $\theta = \tan^{-1}(x^4)$, which implies $\tan \theta = \frac{x^4}{1}$. By considering a right-angled triangle with opposite side $x^4$ and adjacent side $1$, the hypotenuse $h$ is given by $\sqrt{(x^4)^2 + 1^2} = \sqrt{x^8 + 1}$.

Therefore, $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^8 + 1}}$.

Replacing the cosine term with its algebraic equivalent:

$I = -\frac{1}{4} \left( \frac{1}{\sqrt{x^8 + 1}} \right) + C$

Final Solution: $-\frac{1}{4\sqrt{x^8+1}} + C$

Step 1: Problem Definition and Algebraic Manipulation

We are required to evaluate the integral:

$I = \int \frac{x}{\sqrt{x+4}} \, dx$

Given the structure of the integrand, we utilize the algebraic substitution method to simplify the radical expression [Per the Method of Substitution for Integration].

Step 2: Substitution Method

Let $u = x + 4$. Then, differentiating both sides with respect to $x$ gives:

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

From the substitution equation $u = x + 4$, we isolate $x$ to express the numerator in terms of $u$:

$x = u - 4$

Step 3: Transforming the Integral

Substitute $x = u - 4$, $dx = du$, and $\sqrt{x+4} = \sqrt{u}$ into the integral expression:

$I = \int \frac{u - 4}{\sqrt{u}} \, du$

$I = \int \left( \frac{u}{\sqrt{u}} - \frac{4}{\sqrt{u}} \right) \, du$

$I = \int \left( u^{1 - 1/2} - 4u^{-1/2} \right) \, du$

$I = \int \left( u^{1/2} - 4u^{-1/2} \right) \, du$

Step 4: Applying Power Rule of Integration

Applying the power rule $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$ [Per the Fundamental Theorem of Calculus]:

$I = \frac{u^{1/2 + 1}}{1/2 + 1} - 4 \cdot \frac{u^{-1/2 + 1}}{-1/2 + 1} + C$

$I = \frac{u^{3/2}}{3/2} - 4 \cdot \frac{u^{1/2}}{1/2} + C$

$I = \frac{2}{3}u^{3/2} - 8u^{1/2} + C$

Step 5: Back-Substitution

Reverting the variable $u$ back to $x+4$:

$I = \frac{2}{3}(x+4)^{3/2} - 8(x+4)^{1/2} + C$

To simplify further, factor out $2(x+4)^{1/2}$:

$I = 2(x+4)^{1/2} \left[ \frac{1}{3}(x+4) - 4 \right] + C$

$I = 2\sqrt{x+4} \left[ \frac{x+4 - 12}{3} \right] + C$

$I = 2\sqrt{x+4} \left[ \frac{x-8}{3} \right] + C$

$I = \frac{2}{3}(x-8)\sqrt{x+4} + C$

Final Solution: $\frac{2}{3}(x-8)\sqrt{x+4} + C$

Step 1: Identifying the Integral and Substitution Strategy

We are tasked to evaluate the indefinite integral:

$I = \int \frac{x}{9 - 4x^2} \, dx$

To solve this, we employ the method of integration by substitution. We observe that the derivative of the denominator $9 - 4x^2$ is proportional to the numerator $x$.

[Justification: The power rule for differentiation states $\frac{d}{dx}(x^n) = nx^{n-1}$. Here, $\frac{d}{dx}(9 - 4x^2) = -8x$.]

Step 2: Performing the Change of Variables

Let $u = 9 - 4x^2$.

Differentiating both sides with respect to $x$:

$\frac{du}{dx} = -8x \implies du = -8x \, dx$

Rearranging to isolate the terms present in the integral:

$x \, dx = -\frac{1}{8} du$

Step 3: Transforming the Integral

Substitute $u$ and $du$ into the original integral $I$:

$I = \int \frac{1}{u} \cdot \left(-\frac{1}{8}\right) du$

$I = -\frac{1}{8} \int \frac{1}{u} \, du$

[Justification: By the Constant Multiple Rule for integration, $\int k f(u) du = k \int f(u) du$.]

Step 4: Evaluating the Integral

Using the standard integration formula for the reciprocal function:

$\int \frac{1}{u} \, du = \ln|u| + C$

[Justification: The Fundamental Theorem of Calculus implies that the antiderivative of $\frac{1}{x}$ is $\ln|x|$.]

$I = -\frac{1}{8} \ln|u| + C$

Step 5: Back-Substitution

Substitute $u = 9 - 4x^2$ back into the expression:

$I = -\frac{1}{8} \ln|9 - 4x^2| + C$

Visual Representation of the integrand

Final Solution: $-\frac{1}{8} \ln|9 - 4x^2| + C$

Step 1: Identification of the Integral and Substitution Method

We are tasked with evaluating the indefinite integral:

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

Observing the integrand, we note the presence of $\log x$ and its derivative $\frac{1}{x}$, which is implicitly present as the denominator. This structure suggests the Method of Substitution [Per the Substitution Rule for Integration].

Step 2: Defining the Substitution Variable

Let $u = 1 + \log x$.

Differentiating both sides with respect to $x$ yields:

$\frac{du}{dx} = \frac{d}{dx}(1) + \frac{d}{dx}(\log x)$

$\frac{du}{dx} = 0 + \frac{1}{x}$

Therefore, $du = \frac{1}{x} \, dx$ [By the Chain Rule of differentiation].

Step 3: Transformation of the Integral

Substituting $u$ and $du$ into the original expression, the integral transforms as follows:

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

$I = \int u^2 \, du$

Step 4: Integration Using the Power Rule

Applying the standard integration formula $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$ [Per the Fundamental Theorem of Calculus]:

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

$I = \frac{u^3}{3} + C$

Step 5: Back-Substitution

Substitute $u = 1 + \log x$ back into the result to obtain the function in terms of $x$:

$I = \frac{(1 + \log x)^3}{3} + C$

Final Solution: $\frac{(1 + \log x)^3}{3} + C$, where $C$ is the constant of integration.

Step 1: Formal Representation of the Integral

Let the integral be denoted by $I$. The expression is given as:

$I = \int \frac{2\cos x - 3\sin x}{6\cos x + 4\sin x} \, dx$

Step 2: Algebraic Manipulation and Factorization

To facilitate the method of substitution, we factor out common terms in the denominator to identify a potential derivative relationship:

$6\cos x + 4\sin x = 2(3\cos x + 2\sin x)$

Substituting this back into the integral:

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

Step 3: Variable Substitution

Let $u$ be the denominator of the integrand [Per the Substitution Rule]:

$u = 3\cos x + 2\sin x$

Differentiate both sides with respect to $x$ to find $du$:

$\frac{du}{dx} = \frac{d}{dx}(3\cos x + 2\sin x)$

$\frac{du}{dx} = 3(-\sin x) + 2(\cos x) = 2\cos x - 3\sin x$

$du = (2\cos x - 3\sin x) \, dx$

Step 4: Transforming the Integral

Substituting $u$ and $du$ into the expression derived in Step 2:

$I = \frac{1}{2} \int \frac{du}{u}$

By the fundamental rule of integration for logarithmic forms [$\int \frac{1}{u} du = \ln|u| + C$]:

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

Step 5: Back-Substitution

Replace $u$ with the original function of $x$:

$I = \frac{1}{2} \ln|3\cos x + 2\sin x| + C$

Visual Representation of the Function

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

Step 1: Problem Definition and Method Selection

We are required to evaluate the indefinite integral of the exponential function: $I = \int e^{2x+3} \, dx$

Since the exponent is a linear function of $x$ in the form $ax + b$, we utilize the Method of Substitution [Per the Integration by Substitution theorem].

Step 2: Performing Variable Substitution

Let $u = 2x + 3$. To find the differential $du$ in terms of $dx$, we differentiate both sides with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(2x + 3)$

$\frac{du}{dx} = 2 + 0 = 2$

Rearranging the terms to solve for $dx$:

$dx = \frac{du}{2}$

Step 3: Transforming the Integral

Substituting $u$ and $dx$ back into the original integral expression:

$I = \int e^u \cdot \frac{du}{2}$

$I = \frac{1}{2} \int e^u \, du$

Step 4: Integration

Applying the standard integral formula for the exponential function, $\int e^u \, du = e^u + C$, where $C$ is the constant of integration [By the Fundamental Theorem of Calculus]:

$I = \frac{1}{2} e^u + C$

Step 5: Re-substitution

Replace $u$ with the original expression $2x + 3$ to revert the solution to the variable $x$:

$I = \frac{1}{2} e^{2x+3} + C$

Visual Representation of the Exponential Growth

Final Solution: \frac{1}{2} e^{2x+3} + C

Step 1: Identifying the Integral and Substitution Variable

We are tasked to evaluate the indefinite integral:

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

To simplify the integrand, we utilize the method of integration by substitution [Per the Substitution Rule for Integrals]. Let $u$ be the function within the radical:

$u = 1 + 2x^2$

Step 2: Differential Analysis

Differentiate $u$ with respect to $x$ to find the relationship between the differentials $du$ and $dx$:

$\frac{du}{dx} = \frac{d}{dx}(1 + 2x^2)$

$\frac{du}{dx} = 0 + 4x$ [By the Power Rule of differentiation, $\frac{d}{dx}x^n = nx^{n-1}$]

$du = 4x \, dx$

Rearranging to isolate the term present in the original integral ($x \, dx$):

$\frac{du}{4} = x \, dx$

Step 3: Transforming the Integral

Substitute the variables $u$ and $du$ into the original integral $I$:

$I = \int \sqrt{u} \cdot \left( \frac{du}{4} \right)$

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

Step 4: Applying the Power Rule for Integration

Integrate $u^{1/2}$ using the formula $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$, where $n = 1/2$:

$I = \frac{1}{4} \left[ \frac{u^{1/2 + 1}}{1/2 + 1} \right] + C$

$I = \frac{1}{4} \left[ \frac{u^{3/2}}{3/2} \right] + C$

$I = \frac{1}{4} \cdot \frac{2}{3} \cdot u^{3/2} + C$

$I = \frac{1}{6} u^{3/2} + C$

Step 5: Back-Substitution

Replace $u$ with the original expression $1 + 2x^2$ to express the result in terms of $x$:

$I = \frac{1}{6} (1 + 2x^2)^{3/2} + C$

Final Solution: $\frac{1}{6} (1 + 2x^2)^{3/2} + C$

Step 1: Identifying the Integral and Substitution Strategy

The objective is to evaluate the indefinite integral:

$I = \int \frac{x}{e^{x^2}} \, dx$

Rewrite the expression using negative exponents for clarity:

$I = \int x e^{-x^2} \, dx$

[By the properties of exponents where $\frac{1}{a^n} = a^{-n}$]

Step 2: Method of Substitution

To simplify the integrand, we select a substitution $u$ such that its derivative $du$ is proportional to the remaining term in the integral. Let:

$u = x^2$

Differentiating both sides with respect to $x$ yields:

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

$du = 2x \, dx$

Rearranging to isolate the term present in the integral ($x \, dx$):

$\frac{du}{2} = x \, dx$

Step 3: Variable Transformation and Integration

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

$I = \int e^{-u} \cdot \left( \frac{du}{2} \right)$

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

Applying the standard integration formula $\int e^{at} \, dt = \frac{e^{at}}{a} + C$ [where $a = -1$]:

$I = \frac{1}{2} \left( \frac{e^{-u}}{-1} \right) + C$

$I = -\frac{1}{2} e^{-u} + C$

Step 4: Back-Substitution

Revert the variable $u$ to its original expression $u = x^2$:

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

Alternatively, expressing this with a positive exponent:

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

Visual Representation of Function Mapping

Final Solution: $-\frac{1}{2e^{x^2}} + C$ or $-\frac{1}{2}e^{-x^2} + C$

Step 1: Formal Representation of the Integral

We are required to evaluate the indefinite integral denoted as $I$:

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

Step 2: Identification of the Integration Method

Observe the integrand. We identify the function $f(x) = \sin^{-1} x$ and its derivative $f'(x) = \frac{1}{\sqrt{1-x^2}}$. Since the derivative of the numerator (or a part of the integrand) is present as a factor in the expression, we utilize the Method of Substitution [Per the Integration by Substitution Rule].

Step 3: Variable Substitution

Let $u = \sin^{-1} x$.

Differentiating both sides with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\sin^{-1} x)$

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

Rearranging for the differential $dx$:

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

Step 4: Algebraic Transformation of the Integral

Substitute $u$ and $du$ into the original integral $I$:

$I = \int u \, du$

[By the Power Rule of Integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$]

$I = \frac{u^{1+1}}{1+1} + C$

$I = \frac{u^2}{2} + C$

Step 5: Re-substitution and Final Result

Substitute back $u = \sin^{-1} x$ to express the integral in terms of $x$:

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

Visual Representation of the Integration Path

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

Step 1: Establishing the Integral and Substitution Variable

The objective is to evaluate the indefinite integral:

$I = \int x \sqrt{x+2} \, dx$

To simplify the integrand containing the square root, we employ the method of substitution [Per the Substitution Rule for Integrals]. Let:

$u = x + 2$

Differentiating both sides with respect to $x$ yields:

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

Furthermore, we express the term $x$ in terms of $u$ to maintain variable consistency:

$x = u - 2$

Step 2: Substitution and Algebraic Expansion

Substitute $x = (u - 2)$, $\sqrt{x+2} = \sqrt{u}$, and $dx = du$ into the original integral:

$I = \int (u - 2) \sqrt{u} \, du$

$I = \int (u - 2) u^{1/2} \, du$

Applying the distributive property [By the Distributive Law of Real Numbers]:

$I = \int (u^1 \cdot u^{1/2} - 2u^{1/2}) \, du$

$I = \int (u^{3/2} - 2u^{1/2}) \, du$

Step 3: Integration and Power Rule Application

We apply the Power Rule of integration, $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$, where $n \neq -1$ [Per the Fundamental Theorem of Calculus]:

$I = \int u^{3/2} \, du - 2 \int u^{1/2} \, du$

$I = \frac{u^{(3/2)+1}}{(3/2)+1} - 2 \left( \frac{u^{(1/2)+1}}{(1/2)+1} \right) + C$

$I = \frac{u^{5/2}}{5/2} - 2 \left( \frac{u^{3/2}}{3/2} \right) + C$

$I = \frac{2}{5} u^{5/2} - 2 \left( \frac{2}{3} \right) u^{3/2} + C$

$I = \frac{2}{5} u^{5/2} - \frac{4}{3} u^{3/2} + C$

Step 4: Back-Substitution

Replace $u$ with the original variable $(x+2)$:

$I = \frac{2}{5} (x+2)^{5/2} - \frac{4}{3} (x+2)^{3/2} + C$

Step 5: Simplification and Factorization

Factor out common terms to reach the most concise form:

$I = 2(x+2)^{3/2} \left( \frac{x+2}{5} - \frac{2}{3} \right) + C$

$I = 2(x+2)^{3/2} \left( \frac{3(x+2) - 10}{15} \right) + C$

$I = 2(x+2)^{3/2} \left( \frac{3x + 6 - 10}{15} \right) + C$

$I = \frac{2}{15} (x+2)^{3/2} (3x - 4) + C$

Final Solution: $\frac{2}{5} (x+2)^{5/2} - \frac{4}{3} (x+2)^{3/2} + C$ or $\frac{2}{15} (x+2)^{3/2} (3x - 4) + C$

Step 1: Problem Identification and Method Selection

The integral to be evaluated is:
$I = \int \frac{2x}{1+x^2} \, dx$
Upon observation, we identify that the derivative of the denominator $1+x^2$ is exactly $2x$, which is present in the numerator. This implies the Method of Substitution is the most efficient analytical approach [Per the substitution rule for integration].

Step 2: Defining the Substitution Variable

Let $u = 1 + x^2$.
To express the integral entirely in terms of $u$, we must find the differential $du$ with respect to $x$:
$\frac{du}{dx} = \frac{d}{dx}(1 + x^2)$
$\frac{du}{dx} = 0 + 2x = 2x$
Therefore, $du = 2x \, dx$ [By the power rule of differentiation $\frac{d}{dx}x^n = nx^{n-1}$].

Step 3: Variable Transformation

Substitute $u$ and $du$ into the original integral expression:
$I = \int \frac{1}{u} \, du$
Applying the fundamental theorem of logarithmic integration, where $\int \frac{1}{u} \, du = \ln|u| + C$ [Where $C$ is the constant of integration].

Step 4: Algebraic Back-Substitution

Substitute $u = 1 + x^2$ back into the result to obtain the function in terms of the original variable $x$:
$I = \ln|1 + x^2| + C$
Since $1 + x^2$ is always strictly positive for all $x \in \mathbb{R}$, the absolute value notation may be simplified to parentheses:
$I = \ln(1 + x^2) + C$

Visualization: Conceptual Geometry of the Substitution

Final Solution: \int \frac{2x}{1+x^2} \, dx = \ln(1 + x^2) + C

Step 1: Formal Representation of the Integral

To integrate the function $I = \int \frac{\sin x}{1+\cos x} \, dx$, we analyze the structure of the integrand. The denominator is a trigonometric function $f(x) = 1 + \cos x$, and its derivative $f'(x) = -\sin x$ is present in the numerator, adjusted by a constant multiplier [Per the Method of Substitution].

Step 2: Selection of Substitution Variable

Let the denominator be represented by the variable $u$:

$u = 1 + \cos x$

Differentiating both sides with respect to $x$ using the chain rule [By the Chain Rule of Calculus]:

$\frac{du}{dx} = \frac{d}{dx}(1) + \frac{d}{dx}(\cos x)$

$\frac{du}{dx} = 0 - \sin x$

$du = -\sin x \, dx$

Consequently, the numerator term $\sin x \, dx$ can be expressed as:

$\sin x \, dx = -du$

Step 3: Variable Transformation and Integration

Substitute the variables $u$ and $du$ back into the original integral expression:

$I = \int \frac{-du}{u}$

$I = -\int \frac{1}{u} \, du$

Applying the standard integral formula $\int \frac{1}{u} \, du = \ln|u| + C$ [Per the fundamental theorem of integration for natural logarithms]:

$I = -\ln|u| + C$

Step 4: Reversion to Original Variable

Replace $u$ with $1 + \cos x$ to obtain the solution in terms of $x$:

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

Step 5: Trigonometric Simplification (Optional)

Using the logarithmic property $a \ln(b) = \ln(b^a)$ and the identity $1 + \cos x = 2\cos^2(\frac{x}{2})$ [By the Half-Angle Identity], the expression can also be represented as:

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

However, the primary result obtained in Step 4 is the most direct application of the substitution method.

Final Solution: $-\ln|1 + \cos x| + C$

Step 1: Identifying the Integral and Substitution Variable

Given the indefinite integral:

$$I = \int \frac{(\log x)^2}{x} \, dx$$

We observe that the derivative of $\log x$ is $\frac{1}{x}$, which is present as a factor in the integrand. This suggests the Method of Substitution [Per the Substitution Rule for Integration].

Step 2: Defining the Substitution

Let $u = \log x$.

Differentiating both sides with respect to $x$ yields:

$$\frac{du}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$$

Rearranging the differential terms, we get:

$$du = \frac{1}{x} \, dx$$

Step 3: Transforming the Integral

Substitute $u$ and $du$ into the original expression [By the Substitution Theorem]:

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

Substituting the variables defined in Step 2:

$$I = \int u^2 \, du$$

Step 4: Evaluating the Power Integral

Apply the Power Rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$ [Fundamental Theorem of Calculus]:

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

Step 5: Re-substituting the Original Variable

Since the original variable was $x$, replace $u$ with $\log x$:

$$I = \frac{(\log x)^3}{3} + C$$

Visualization of the Integral Curve

The following SVG represents the geometric behavior of the derivative $\frac{(\log x)^2}{x}$ for $x > 0$.

Final Solution: \frac{(\log x)^3}{3} + C

Step 1: Formal Representation of the Integral

We are required to evaluate the indefinite integral denoted by $I$: $$I = \int \frac{1}{x(\log x)^m} \, dx$$ [Given the domain constraints $x > 0$ and $m \neq 1$ to ensure the function is defined and the integral does not result in a logarithmic singularity.]

Step 2: Substitution Methodology

To simplify the integrand, we perform a change of variable. Let:

• $u = \log x$
• By applying the differential operator, we obtain: $\frac{du}{dx} = \frac{1}{x}$
• Rearranging for the differential element: $du = \frac{1}{x} \, dx$

[By the Method of Substitution, where we replace a composite function part with a single variable to simplify the integrand.]

Step 3: Transforming the Integral

Substituting $u = \log x$ and $du = \frac{1}{x} \, dx$ into the original integral expression:

$$I = \int \frac{1}{u^m} \, du$$

Rewriting the integrand using the laws of exponents: $$I = \int u^{-m} \, du$$

Step 4: Integration Using the Power Rule

Applying the standard integration power rule, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ (where $n \neq -1$):

$$I = \frac{u^{-m+1}}{-m+1} + C$$

Simplifying the exponent and the denominator: $$I = \frac{u^{1-m}}{1-m} + C$$

Step 5: Re-substitution and Final Expression

Substitute $u = \log x$ back into the expression to return to the original variable $x$:

$$I = \frac{(\log x)^{1-m}}{1-m} + C$$

Final Solution: \frac{(\log x)^{1-m}}{1-m} + C

Step 1: Identifying the Integral and Substitution Variable

We are tasked with solving the indefinite integral:

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

To simplify the expression, we apply the Method of Substitution [By the Change of Variables Theorem]. Let us define a new variable $u$ such that its derivative is present within the integrand.

Let $u = \tan^{-1} x$.

Step 2: Differentiating and Substitution

We compute the differential $du$ with respect to $x$ [Per the Chain Rule of Calculus]:

$\frac{du}{dx} = \frac{d}{dx}(\tan^{-1} x)$

$\frac{du}{dx} = \frac{1}{1+x^2}$

$du = \frac{1}{1+x^2} \, dx$

Step 3: Transforming the Integral

Substitute the expressions for $u$ and $du$ back into the original integral $I$:

$I = \int e^{u} \, du$

Step 4: Performing Integration

Integrating $e^u$ with respect to $u$ follows the standard rule for exponential functions [Per the Fundamental Theorem of Calculus]:

$I = e^u + C$

Where $C$ represents the constant of integration.

Step 5: Back-Substitution

We replace the variable $u$ with the original function of $x$ to express the solution in terms of the initial independent variable:

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

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

Step 1: Problem Definition and Substitution Strategy

The objective is to evaluate the indefinite integral:

$I = \int \cot x \log(\sin x) \, dx$

To simplify this integral, we observe the relationship between the logarithmic function and its derivative. Let $u = \log(\sin x)$.

[By the Chain Rule of differentiation]:

$\frac{du}{dx} = \frac{d}{dx} (\log(\sin x)) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{\cos x}{\sin x} = \cot x$

Step 2: Differential Transformation

From the substitution $u = \log(\sin x)$, we derive the differential form:

$du = \cot x \, dx$

Substituting these components into the original integral $I$:

$I = \int (\log(\sin x)) (\cot x \, dx)$

$I = \int u \, du$

Step 3: Integration and Back-Substitution

Applying the Power Rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$:

$\int u^1 \, du = \frac{u^{1+1}}{1+1} + C = \frac{u^2}{2} + C$

[Applying the Fundamental Theorem of Calculus for indefinite integrals].

Now, replace $u$ with the original variable $\log(\sin x)$:

$I = \frac{(\log(\sin x))^2}{2} + C$

Visual Representation of the Substitution Logic

Step 4: Conclusion

The integral of the function $\cot x \log(\sin x)$ with respect to $x$ is obtained by identifying the function and its derivative present within the integrand.

Final Solution: \frac{1}{2} (\log(\sin x))^2 + C

Given Integral

We are required to evaluate the indefinite integral:

$I = \int \sin x \sin(\cos x) \, dx$

Step 1: Identification of the Substitution Method

To solve this integral, we observe the composition of functions within the integrand. We have $\sin x$ multiplied by $\sin(\cos x)$. Notice that the derivative of the inner function $\cos x$ is $-\sin x$, which is present in the integrand. Therefore, we employ the Method of Substitution [Per the Integration by Substitution rule].

Step 2: Performing the Substitution

Let the inner function be defined as $u$:

$u = \cos x$

Now, differentiate $u$ with respect to $x$ to find the differential $du$:

$\frac{du}{dx} = -\sin x$

Rearranging for the differential components present in the original integral:

$du = -\sin x \, dx \implies -du = \sin x \, dx$

Step 3: Transforming the Integral

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

$I = \int \sin(\cos x) \cdot (\sin x \, dx)$

$I = \int \sin(u) \cdot (-du)$

$I = -\int \sin(u) \, du$

Step 4: Evaluating the Integral

Utilizing the standard integral formula for the sine function, $\int \sin(u) \, du = -\cos(u) + C$ [By fundamental integration axioms]:

$I = -(-\cos(u)) + C$

$I = \cos(u) + C$

Step 5: Back-Substitution

Substitute the original variable $x$ back into the expression by replacing $u$ with $\cos x$:

$I = \cos(\cos x) + C$

Visual Representation of the Differential Substitution

Final Solution: $\cos(\cos x) + C$, where $C$ is the constant of integration.

Step 1: Problem Definition and Algebraic Simplification

We are tasked with evaluating the indefinite integral:

$I = \int \sin(ax + b) \cos(ax + b) \, dx$

To simplify the integrand, we utilize the trigonometric identity for the sine of a double angle: $\sin(2\theta) = 2 \sin \theta \cos \theta$. Rearranging this identity gives:

$\sin \theta \cos \theta = \frac{1}{2} \sin(2\theta)$

Substituting $\theta = ax + b$ into the identity:

$I = \int \frac{1}{2} \sin(2(ax + b)) \, dx = \frac{1}{2} \int \sin(2ax + 2b) \, dx$ [By Trigonometric Double-Angle Identity]

Step 2: Integration via Substitution Method

Let $u = 2ax + 2b$. To find the differential $du$, we differentiate with respect to $x$:

$\frac{du}{dx} = 2a \implies dx = \frac{du}{2a}$

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

$I = \frac{1}{2} \int \sin(u) \cdot \frac{du}{2a}$

$I = \frac{1}{4a} \int \sin(u) \, du$

Step 3: Evaluating the Integral

Applying the standard integral formula for the sine function, $\int \sin(u) \, du = -\cos(u) + C$:

$I = \frac{1}{4a} (-\cos(u)) + C$

Re-substituting $u = 2ax + 2b$ back into the expression:

$I = -\frac{\cos(2ax + 2b)}{4a} + C$

Step 4: Alternative Analytical Verification

Alternatively, using the method of substitution $t = \sin(ax + b)$:

$dt = a \cos(ax + b) \, dx \implies \cos(ax + b) \, dx = \frac{dt}{a}$

$I = \int t \cdot \frac{dt}{a} = \frac{1}{a} \cdot \frac{t^2}{2} + C = \frac{\sin^2(ax + b)}{2a} + C$

Note: These two forms are equivalent up to a constant, as $\cos(2\theta) = 1 - 2\sin^2\theta$.

Final Solution: $-\frac{\cos(2ax + 2b)}{4a} + C$ or $\frac{\sin^2(ax + b)}{2a} + C$