Step 1: Application of Integration by Parts

To evaluate the integral $I = \int x \cos^{-1} x \, dx$, we employ the method of Integration by Parts, which states: $\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \cdot \int v \, dx \right) dx$ [Per the Integration by Parts Theorem].

According to the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), we assign:

• $u = \cos^{-1} x$
• $dv = x \, dx$

Differentiating $u$ and integrating $dv$ yields:

• $\frac{du}{dx} = -\frac{1}{\sqrt{1-x^2}}$
• $v = \int x \, dx = \frac{x^2}{2}$

Step 2: Substitution into the Formula

Applying the components to the integration formula:

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

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

Step 3: Evaluating the Remaining Integral

To integrate $\int \frac{x^2}{\sqrt{1-x^2}} \, dx$, we manipulate the numerator to create a term that matches the denominator:

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

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

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

Step 4: Applying Standard Integral Formulas

We use the following standard results:

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

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

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

Step 5: Final Assembly

Substitute the result back into the expression for $I$:

$I = \frac{x^2}{2} \cos^{-1} x + \frac{1}{2} \left[ -\frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x \right] + C$

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

Final Solution: $\frac{x^2}{2} \cos^{-1} x - \frac{x}{4}\sqrt{1-x^2} + \frac{1}{4}\sin^{-1} x + C$

Step 1: Manipulating the Integrand

Given the integral: $I = \int \frac{(x-3)e^x}{(x-1)^3} \, dx$.

To simplify the integrand, we express the numerator $(x-3)$ in terms of $(x-1)$ to facilitate partial fraction decomposition relative to the denominator $(x-1)^3$.

Rewrite $(x-3)$ as $(x-1) - 2$.

Substituting this into the integral:

$I = \int \frac{((x-1) - 2)e^x}{(x-1)^3} \, dx$
$I = \int \left( \frac{x-1}{(x-1)^3} - \frac{2}{(x-1)^3} \right) e^x \, dx$
$I = \int \left( \frac{1}{(x-1)^2} - \frac{2}{(x-1)^3} \right) e^x \, dx$ [By Algebraic Simplification]

Step 2: Applying the Integration Formula

We utilize the standard integration identity: $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$.

Let $f(x) = \frac{1}{(x-1)^2}$.

Differentiating $f(x)$ with respect to $x$ using the Power Rule and Chain Rule:

$f'(x) = \frac{d}{dx} (x-1)^{-2}$
$f'(x) = -2(x-1)^{-3} \cdot \frac{d}{dx}(x-1)$ [Per the Chain Rule]
$f'(x) = \frac{-2}{(x-1)^3}$

Step 3: Verification of the Identity

Our integrand is now in the form $e^x [f(x) + f'(x)]$, where:

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

Substituting these back into the integral expression:

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

Step 4: Final Integration

Applying the formula $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$:

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

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

Given Integral

To evaluate the integral: $I = \int x \tan^{-1} x \, dx$

Step 1: Selection of Integration by Parts Technique

We employ the method of Integration by Parts, defined by the formula: $\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx$ [Per the Integration by Parts theorem].

To determine $u$ and $v$, we utilize the ILATE rule (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential). Here, $u = \tan^{-1} x$ (Inverse) and $v = x$ (Algebraic).

Step 2: Assigning Components

• $u = \tan^{-1} x \implies \frac{du}{dx} = \frac{1}{1 + x^2}$ [By derivative of inverse tangent function]
• $\int v \, dx = \int x \, dx = \frac{x^2}{2}$

Step 3: Application of the Formula

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

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

Step 4: Algebraic Manipulation of the Remaining Integral

To integrate $\frac{x^2}{1 + x^2}$, perform polynomial adjustment: $x^2 = (x^2 + 1) - 1$.

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

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

$= x - \tan^{-1} x$ [By standard integral $\int \frac{1}{1+x^2} dx = \tan^{-1} x + C$]

Step 5: Consolidation of Terms

Substituting the result from Step 4 back into the expression from Step 3:

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

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

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

Visual Representation of Function Area

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

Step 1: Application of Integration by Parts

To evaluate the integral $I = \int (x^2+1) \log x \, dx$, we employ the technique of Integration by Parts, defined by the formula:

$\int u(x)v(x) \, dx = u(x) \int v(x) \, dx - \int \left( \frac{du}{dx} \int v(x) \, dx \right) dx$ [Per the Integration by Parts Rule]

Following the LIATE rule (Logarithmic, Inverse Trigonometric, Algebraic, Trigonometric, Exponential), we assign the variables:

• $u = \log x \implies \frac{du}{dx} = \frac{1}{x}$
• $v = x^2 + 1 \implies \int v \, dx = \frac{x^3}{3} + x$

Step 2: Execution of the Integration Formula

Substituting the assigned variables into the Integration by Parts formula:

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

Step 3: Simplification of the Resulting Integral

Distribute $\frac{1}{x}$ into the term within the integrand:

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

Apply the linearity property of integration [Per the Sum Rule of Calculus]:

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

Step 4: Evaluating the Remaining Integral

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

$I = \left( \frac{x^3}{3} + x \right) \log x - \left( \frac{x^3}{3 \cdot 3} + x \right) + C$

$I = \left( \frac{x^3}{3} + x \right) \log x - \left( \frac{x^3}{9} + x \right) + C$

Final Solution

Factorizing and presenting the final expression in its most concise mathematical form:

Final Solution: $\left( \frac{x^3}{3} + x \right) \log x - \frac{x^3}{9} - x + C$

Step 1: Analyzing the Integral Structure

The given integral is:
$I = \int e^x \sec x (1 + \tan x) dx$

Distribute the $\sec x$ term within the parenthesis:
$I = \int e^x (\sec x + \sec x \tan x) dx$

Step 2: Identifying the Relevant Integral Identity

We utilize the standard integration identity involving exponential functions:
$\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$ [Per the Fundamental Theorem of Calculus and the rule for integrating exponential functions of the form $e^x(f(x) + f'(x))$].

Step 3: Verification of Functions

Let $f(x) = \sec x$.
To verify the identity, compute the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(\sec x)$
$f'(x) = \sec x \tan x$ [By the derivative rule for trigonometric functions].

Since the integrand matches the form $e^x [f(x) + f'(x)]$, where $f(x) = \sec x$ and $f'(x) = \sec x \tan x$, the conditions of the identity are satisfied.

Step 4: Applying the Integral Formula

Substitute $f(x) = \sec x$ into the result of the identity $e^x f(x) + C$:
$I = e^x \sec x + C$

Visual Representation of the Exponential Growth and Function

Conclusion

Comparing the result $e^x \sec x + C$ with the standard options provided in the examination context, this solution corresponds precisely to the evaluated integral.

Final Solution: $e^x \sec x + C$

1. Theoretical Foundation: Integration by Parts

To evaluate the integral of the product of two functions, we employ the Integration by Parts formula: $$\int u(x)v(x) \, dx = u(x) \int v(x) \, dx - \int \left( \frac{du}{dx} \cdot \int v(x) \, dx \right) dx$$ [Per the Integration by Parts theorem]. To determine the choice of $u$ and $v$, we adhere to the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential). In the expression $x^2 e^x$, $x^2$ is an algebraic function and $e^x$ is an exponential function.

2. Application of LIATE and Initial Setup

Following LIATE, we set:

• $u = x^2$
• $dv = e^x \, dx$

Differentiating and integrating respectively:

• $\frac{du}{dx} = 2x \implies du = 2x \, dx$
• $v = \int e^x \, dx = e^x$

3. First Iteration of Integration by Parts

Substituting these values into the formula: $$\int x^2 e^x \, dx = x^2 e^x - \int (e^x \cdot 2x) \, dx$$ $$\int x^2 e^x \, dx = x^2 e^x - 2 \int x e^x \, dx$$

4. Second Iteration (Evaluating $\int x e^x \, dx$)

We apply Integration by Parts again for the remaining integral $\int x e^x \, dx$:

• Let $u_1 = x \implies du_1 = dx$
• Let $dv_1 = e^x \, dx \implies v_1 = e^x$

Substituting into the formula: $$\int x e^x \, dx = x e^x - \int e^x \, dx$$ $$\int x e^x \, dx = x e^x - e^x + C_1$$

5. Final Synthesis

Substitute the result of the second integral back into the expression from Step 3: $$\int x^2 e^x \, dx = x^2 e^x - 2(x e^x - e^x) + C$$ Expanding the brackets: $$\int x^2 e^x \, dx = x^2 e^x - 2x e^x + 2e^x + C$$ Factor out the common term $e^x$: $$\int x^2 e^x \, dx = e^x(x^2 - 2x + 2) + C$$

Visualization of the Integral Growth

Final Solution: $e^x(x^2 - 2x + 2) + C$

Step 1: Identifying the Integration Technique

To evaluate the integral of $I = \int \tan^{-1} x \, dx$, we employ the method of Integration by Parts [Per the Integration by Parts formula: $\int u \, dv = uv - \int v \, du$]. Since $\tan^{-1} x$ does not have a standard elementary integral, we treat the integrand as a product of two functions: $f(x) = \tan^{-1} x$ and $g(x) = 1$.

Step 2: Assigning Components (LIATE Rule)

Following the LIATE rule (Logarithmic, Inverse Trigonometric, Algebraic, Trigonometric, Exponential), we assign:

• $u = \tan^{-1} x$
• $dv = 1 \, dx$

Differentiating $u$ with respect to $x$:
$\frac{du}{dx} = \frac{1}{1 + x^2} \implies du = \frac{1}{1 + x^2} \, dx$ [Standard derivative of inverse trigonometric functions]

Integrating $dv$:
$\int dv = \int 1 \, dx \implies v = x$

Step 3: Applying the Integration by Parts Formula

Substitute the assigned values into the formula $\int u \, dv = uv - \int v \, du$:

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

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

Step 4: Solving the Remaining Integral

To solve $\int \frac{x}{1 + x^2} \, dx$, use the method of substitution. Let $t = 1 + x^2$.

Then, $\frac{dt}{dx} = 2x \implies dt = 2x \, dx \implies x \, dx = \frac{dt}{2}$.

Substituting into the integral:

$\int \frac{x}{1 + x^2} \, dx = \int \frac{1}{t} \cdot \frac{dt}{2} = \frac{1}{2} \ln |t| + C_1$

Substituting $t = 1 + x^2$ back into the expression:

$\int \frac{x}{1 + x^2} \, dx = \frac{1}{2} \ln(1 + x^2) + C_1$

Step 5: Final Assembly

Combining the terms obtained in Step 3 and Step 4:

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

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

Step 1: Application of Integration by Parts

To evaluate the integral $I = \int x (\log x)^2 \, dx$, we utilize the technique of integration by parts. The formula for integration by parts is defined as:

$\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx$

Following the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), we assign the functions as follows:

• Let $u = (\log x)^2$ [The logarithmic function takes precedence over the algebraic function].
• Let $dv = x \, dx$.

Step 2: Differentiation and Integration of Assigned Parts

We compute the derivative of $u$ and the integral of $dv$:

• $\frac{du}{dx} = 2(\log x) \cdot \frac{d}{dx}(\log x) = 2(\log x) \cdot \frac{1}{x} = \frac{2 \log x}{x}$ [By the Chain Rule of differentiation].
• $v = \int x \, dx = \frac{x^2}{2}$ [By the Power Rule of integration: $\int x^n dx = \frac{x^{n+1}}{n+1}$].

Step 3: Execution of the Integration by Parts Formula

Substituting the values into the formula:

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

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

Step 4: Evaluating the Secondary Integral $\int x \log x \, dx$

We apply integration by parts again for $\int x \log x \, dx$:

• Let $u = \log x \implies \frac{du}{dx} = \frac{1}{x}$
• Let $dv = x \, dx \implies v = \frac{x^2}{2}$

Applying the formula:

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

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

$\int x \log x \, dx = \frac{x^2 \log x}{2} - \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2 \log x}{2} - \frac{x^2}{4}$

Step 5: Synthesizing the Final Expression

Substituting the result from Step 4 back into the equation for $I$:

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

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

Geometrical Representation of the Integral Area

Final Solution: $\frac{x^2}{2} (\log x)^2 - \frac{x^2}{2} \log x + \frac{x^2}{4} + C$

Step 1: Identifying the Integration Technique

To evaluate the integral $I = \int x \sin 3x \, dx$, we employ the method of Integration by Parts. This method is defined by the formula:

$\int u(x) v(x) \, dx = u(x) \int v(x) \, dx - \int \left( \frac{du}{dx} \int v(x) \, dx \right) \, dx$ [Per the Integration by Parts Rule]

Step 2: Selecting Functions (ILATE Rule)

Following the ILATE hierarchy (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential), we assign the functions:

• $u = x$ (Algebraic function)
• $dv = \sin 3x \, dx$ (Trigonometric function)

Calculating the derivatives and integrals:

• $\frac{du}{dx} = 1 \implies du = dx$
• $v = \int \sin 3x \, dx = -\frac{\cos 3x}{3}$ [Per the integration rule $\int \sin(ax) \, dx = -\frac{\cos(ax)}{a} + C$]

Step 3: Executing the Integration by Parts Formula

Substitute $u, v, du,$ and $dv$ into the formula:

$I = x \left( -\frac{\cos 3x}{3} \right) - \int \left( 1 \cdot -\frac{\cos 3x}{3} \right) \, dx$

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

Step 4: Final Evaluation of the Integral

Integrate $\cos 3x$:

$\int \cos 3x \, dx = \frac{\sin 3x}{3} + C$ [Per the integration rule $\int \cos(ax) \, dx = \frac{\sin(ax)}{a} + C$]

Substituting this back into the expression for $I$:

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

$I = -\frac{x \cos 3x}{3} + \frac{\sin 3x}{9} + C$

Visual Representation of the Integrand Area

Final Solution: $-\frac{x \cos 3x}{3} + \frac{\sin 3x}{9} + C$

Initial Setup: Identifying the Method of Integration

To evaluate the integral $I = \int x \sec^2 x \, dx$, we apply the method of Integration by Parts. The formula for Integration by Parts, derived from the product rule of differentiation, is given by:

$\int u(x) v'(x) \, dx = u(x)v(x) - \int u'(x) v(x) \, dx$ [Per the Integration by Parts Theorem]

We select $u(x)$ and $v'(x)$ based on the ILATE rule (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential), which dictates the hierarchy for choosing $u(x)$.

Step 1: Assigning Variables

• Let $u = x$ (Algebraic function). [Choosing $u$ to be the function that simplifies upon differentiation].
• Let $dv = \sec^2 x \, dx$ (Trigonometric function).
• Then, differentiate $u$: $du = dx$.
• Then, integrate $dv$: $v = \int \sec^2 x \, dx = \tan x$. [Using the standard integral formula $\int \sec^2 x \, dx = \tan x + C$].

Step 2: Applying the Integration by Parts Formula

Substitute the identified components into the integration formula:

$I = x \tan x - \int \tan x \, dx$

Step 3: Evaluating the Remaining Integral

The integral of $\tan x$ is a standard integral derived from the substitution $u = \cos x$:

$\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx = -\ln|\cos x| + C = \ln|\sec x| + C$ [Using the property $\ln(a^{-1}) = -\ln(a)$].

Step 4: Final Assembly

Substitute the result of the integral back into the expression for $I$:

$I = x \tan x - (-\ln|\cos x|) + C$

$I = x \tan x + \ln|\cos x| + C$

Visual Representation of the Functions

Final Solution: $x \tan x + \ln|\sec x| + C$

Given Variables & Initial Setup

To evaluate the integral of the product of two functions, we employ the method of Integration by Parts. The integral is defined as:

$I = \int x \sin x \, dx$

The standard formula for Integration by Parts is given by: [Per the Integration by Parts Rule]

$\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \cdot \int v \, dx \right) dx$

Step 1: Selecting $u$ and $v$ using the ILATE Rule

According to the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential), algebraic functions take precedence over trigonometric functions for the choice of $u$.

• Let $u = x$ (Algebraic function)
• Let $dv = \sin x \, dx$ (Trigonometric function)

Differentiating $u$ with respect to $x$: $\frac{du}{dx} = 1 \implies du = dx$

Integrating $dv$: $\int dv = \int \sin x \, dx \implies v = -\cos x$

Step 2: Applying the Integration by Parts Formula

Substitute the identified components into the integration formula:

$I = (x)(-\cos x) - \int (1)(-\cos x) \, dx$

$I = -x \cos x + \int \cos x \, dx$

Step 3: Evaluating the Final Integral

Evaluate the remaining integral $\int \cos x \, dx$, which yields $\sin x$: [By the Fundamental Theorem of Calculus]

$I = -x \cos x + \sin x + C$

Where $C$ represents the constant of integration.

Visual Representation of the Functions

Final Solution: $-x \cos x + \sin x + C$

Step 1: Analyzing the Integrand

The given integral is: $I = \int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) \, dx$ [Per the general form $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$]

Step 2: Trigonometric Simplification

We simplify the expression $\frac{1 + \sin x}{1 + \cos x}$ using half-angle identities:

• $1 + \cos x = 2 \cos^2 \left( \frac{x}{2} \right)$ [Using $\cos 2\theta = 2\cos^2 \theta - 1$]
• $\sin x = 2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)$ [Using $\sin 2\theta = 2 \sin \theta \cos \theta$]

Substituting these into the expression:

$f(x) = \frac{1 + 2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)}$

Step 3: Algebraic Decomposition

Divide the terms in the numerator by the denominator:

$f(x) = \frac{1}{2 \cos^2(x/2)} + \frac{2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)}$ $f(x) = \frac{1}{2} \sec^2\left( \frac{x}{2} \right) + \tan\left( \frac{x}{2} \right)$

Step 4: Identifying the Function and its Derivative

Let $f(x) = \tan\left( \frac{x}{2} \right)$.

Then, the derivative $f'(x)$ is:

$f'(x) = \frac{d}{dx} \left[ \tan\left( \frac{x}{2} \right) \right] = \sec^2\left( \frac{x}{2} \right) \cdot \frac{1}{2}$ $f'(x) = \frac{1}{2} \sec^2\left( \frac{x}{2} \right)$

This matches our decomposed term, confirming the structure $e^x [f(x) + f'(x)]$.

Step 5: Applying the Integral Formula

Using the standard integration result for exponential-trigonometric products:

$\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$

Substituting $f(x) = \tan\left( \frac{x}{2} \right)$:

$I = e^x \tan\left( \frac{x}{2} \right) + C$

Final Solution: $e^x \tan\left( \frac{x}{2} \right) + C$

Given Integral

We are required to evaluate the indefinite integral:

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

Step 1: Identifying the Mathematical Form

The integrand is of the form $e^x [f(x) + f'(x)]$. We utilize the standard integration identity derived from the product rule of differentiation [By the Theorem of Integration by Parts]:

$\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$

Step 2: Defining the Functions

Let $f(x) = \frac{1}{x}$.

To verify the existence of the derivative component, we differentiate $f(x)$ with respect to $x$ using the Power Rule $\frac{d}{dx}(x^n) = nx^{n-1}$:

$f'(x) = \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2} = -\frac{1}{x^2}$

Since the derivative of $\frac{1}{x}$ is exactly $-\frac{1}{x^2}$, our integrand fits the required form perfectly.

Step 3: Application of the Integration Identity

Substituting $f(x) = \frac{1}{x}$ and $f'(x) = -\frac{1}{x^2}$ into the identity $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$, we obtain:

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

$I = e^x \cdot \left( \frac{1}{x} \right) + C$

Visual Representation of Function Relationship

Step 4: Final Verification

Differentiating the result to ensure precision [By the Fundamental Theorem of Calculus]:

$\frac{d}{dx} \left( \frac{e^x}{x} + C \right) = \frac{x \cdot \frac{d}{dx}(e^x) - e^x \cdot \frac{d}{dx}(x)}{x^2} + 0$ [Using the Quotient Rule $\frac{d}{dx}(\frac{u}{v}) = \frac{vu' - uv'}{v^2}$]

$= \frac{x e^x - e^x}{x^2} = \frac{e^x(x - 1)}{x^2} = e^x \left( \frac{x}{x^2} - \frac{1}{x^2} \right) = e^x \left( \frac{1}{x} - \frac{1}{x^2} \right)$

The derivative confirms our result is accurate.

Final Solution: $\frac{e^x}{x} + C$

Initial Setup: Integration by Parts

To evaluate the integral $I = \int x \sin^{-1} x \, dx$, we employ the method of Integration by Parts. The formula for Integration by Parts is given by:

$\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx$ [Per the LIATE rule for selection of $u$ and $v$]

According to the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), we assign:

• $u = \sin^{-1} x$ (Inverse trigonometric function)
• $dv = x \, dx$ (Algebraic function)

Step 1: Calculating Derivatives and Integrals

Differentiating $u$ with respect to $x$:

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

Integrating $dv$ with respect to $x$:

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

Step 2: Applying the Integration by Parts Formula

Substituting the values into the formula:

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

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

Step 3: Solving the Remaining Integral

To solve $\int \frac{x^2}{\sqrt{1 - x^2}} \, dx$, we manipulate the numerator:

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

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

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

Using the standard integration formulas:

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

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

Thus:

$\int \frac{x^2}{\sqrt{1 - x^2}} \, dx = -\left[ \frac{x}{2} \sqrt{1 - x^2} + \frac{1}{2} \sin^{-1} x \right] + \sin^{-1} x$

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

Step 4: Final Assembly

Substitute the result back into the main equation $I$:

$I = \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} \left[ -\frac{x}{2} \sqrt{1 - x^2} + \frac{1}{2} \sin^{-1} x \right] + C$

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

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

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

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

Given Integral

To evaluate the integral: $I = \int x^2 \log x \, dx$

Step 1: Application of Integration by Parts

The method of Integration by Parts is defined by the formula: $\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \cdot \int v \, dx \right) dx$ [By the Product Rule of Integration]

To choose $u$ and $v$, we utilize the ILATE rule (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential). Here, $\log x$ is a logarithmic function and $x^2$ is an algebraic function.

• Let $u = \log x$
• Let $v = x^2$

Step 2: Differentiation and Integration of Selected Functions

Calculating the derivatives and integrals required for the formula:

• $\frac{du}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$
• $\int v \, dx = \int x^2 \, dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3}$ [Using the Power Rule for Integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1}$]

Step 3: Substitution into the Parts Formula

Substituting the values into the integration by parts formula:

$I = (\log x) \cdot \left( \frac{x^3}{3} \right) - \int \left( \frac{1}{x} \cdot \frac{x^3}{3} \right) dx$

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

Step 4: Final Integration

Now, compute the remaining integral:

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

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

$I = \frac{x^3}{3} \log x - \frac{x^3}{9} + C$

Visual Representation of Function Growth

Factor out $\frac{x^3}{3}$ to simplify the expression:

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

Final Solution: $\frac{x^3}{3} \log x - \frac{x^3}{9} + C$

Given Integral

Evaluate the integral: $I = \int \frac{x e^x}{(1+x)^2} \, dx$

Step 1: Algebraic Manipulation of the Numerator

To simplify the integrand, express the numerator $x$ in terms of the denominator $(1+x)$. This facilitates splitting the fraction into simpler integrable components.

$x = (x + 1) - 1$

Substitute this into the original expression:

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

[By algebraic property of addition and subtraction]

Step 2: Splitting the Integrand

Distribute the terms in the numerator over the denominator $(1+x)^2$:

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

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

Step 3: Applying the Standard Integral Identity

Recall the standard integration formula: $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$

[Per the property of Exponential Integrals]

Let $f(x) = \frac{1}{1+x}$

Find the derivative $f'(x)$ using the Power Rule and Chain Rule:

$f'(x) = \frac{d}{dx} (1+x)^{-1} = -1(1+x)^{-2} \cdot \frac{d}{dx}(1+x)$

$f'(x) = -\frac{1}{(1+x)^2}$

Step 4: Final Integration

Observe that the integrand is exactly in the form $e^x [f(x) + f'(x)]$.

Substituting $f(x) = \frac{1}{1+x}$ and $f'(x) = -\frac{1}{(1+x)^2}$ into the identity:

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

Visual Representation of the Function

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

Step 1: Application of Integration by Parts Formula

To evaluate the integral $\int x \log(2x) \, dx$, we utilize the method of Integration by Parts, which states: $\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \cdot \int v \, dx \right) dx$ [Integration by Parts Formula]

According to the ILATE rule (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential), we assign the functions as follows:

• Let $u = \log(2x)$ [Logarithmic function chosen as the first function]
• Let $dv = x \, dx$ [Algebraic function chosen as the second function]

Step 2: Differentiation and Integration of Assigned Parts

Calculate the derivative of $u$ and the integral of $dv$:

• $\frac{du}{dx} = \frac{d}{dx}(\log(2x)) = \frac{1}{2x} \cdot 2 = \frac{1}{x}$ [Chain Rule of Differentiation]
• $v = \int x \, dx = \frac{x^2}{2}$ [Power Rule of Integration: $\int x^n dx = \frac{x^{n+1}}{n+1}$]

Step 3: Substitution into the Formula

Substitute these components into the Integration by Parts formula: $\int x \log(2x) \, dx = \left( \log(2x) \cdot \frac{x^2}{2} \right) - \int \left( \frac{1}{x} \cdot \frac{x^2}{2} \right) dx$

Step 4: Algebraic Simplification and Final Integration

Simplify the integrand within the integral sign: $\int x \log(2x) \, dx = \frac{x^2}{2} \log(2x) - \int \frac{x}{2} \, dx$ $\int x \log(2x) \, dx = \frac{x^2}{2} \log(2x) - \frac{1}{2} \int x \, dx$

Perform the final integration: $= \frac{x^2}{2} \log(2x) - \frac{1}{2} \left( \frac{x^2}{2} \right) + C$ [Where $C$ is the constant of integration]

Step 5: Final Expression Consolidation

Combining the terms: $= \frac{x^2}{2} \log(2x) - \frac{x^2}{4} + C$

Final Solution: $\frac{x^2}{2} \log(2x) - \frac{x^2}{4} + C$

Step 1: Analyzing the Integral Structure

The given integral is $I = \int x^2 e^{x^3} \, dx$. We observe that the exponent of the exponential function $e$ is $x^3$, and the derivative of $x^3$ with respect to $x$ is $3x^2$. Since the term $x^2$ is present in the integrand, this integral is ideally suited for the Method of Substitution [Ref: Fundamental Theorem of Calculus and Substitution Rule].

Step 2: Performing Variable Substitution

Let $u = x^3$.

Differentiating both sides with respect to $x$:

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

$du = 3x^2 \, dx$

Rearranging to isolate the terms present in the original integrand:

$\frac{1}{3} \, du = x^2 \, dx$

Step 3: Transforming the Integral

Substituting $u$ and $du$ back into the integral $I$:

$I = \int e^u \cdot \frac{1}{3} \, du$

$I = \frac{1}{3} \int e^u \, du$

Step 4: Integration and Back-Substitution

Applying the standard integral rule for the exponential function $\int e^u \, du = e^u + C$ [Ref: Integration Rules of Transcendental Functions]:

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

Substituting back $u = x^3$:

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

Visual Representation of Substitution Geometry

Summary of Calculation

Original Form	Substitution	Resultant Form
$\int x^2 e^{x^3} \, dx$	$u = x^3, du = 3x^2 \, dx$	$\frac{1}{3} e^u + C$

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

Step 1: Theoretical Foundation

To evaluate the integral $I = \int e^x (\sin x + \cos x) \, dx$, we utilize the standard integral property related to the exponential function: $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$ [Per the property of Integration by Parts where $f(x)$ is a differentiable function].

Step 2: Decomposition of the Integrand

Let $f(x) = \sin x$. We compute the derivative of $f(x)$ with respect to $x$: $f'(x) = \frac{d}{dx}(\sin x) = \cos x$. [By standard derivative formula for trigonometric functions].

Step 3: Application of the Integration Formula

Substituting $f(x) = \sin x$ and $f'(x) = \cos x$ into the structure $e^x [f(x) + f'(x)]$: $I = \int e^x (\sin x + \cos x) \, dx$ Applying the theorem $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$: $I = e^x \sin x + C$

Step 4: Verification via Integration by Parts (Manual Derivation)

We can verify the result using $\int u \, dv = uv - \int v \, du$. Let $I = \int e^x \sin x \, dx + \int e^x \cos x \, dx$. Evaluate $I_1 = \int e^x \sin x \, dx$ using Parts: $u = \sin x, dv = e^x dx \implies du = \cos x dx, v = e^x$. $I_1 = e^x \sin x - \int e^x \cos x \, dx$. Substituting $I_1$ back into the original expression: $I = (e^x \sin x - \int e^x \cos x \, dx) + \int e^x \cos x \, dx$. $I = e^x \sin x + C$.

Visual Representation of the Function

Final Solution: $e^x \sin x + C$

Step 1: Application of Integration by Parts

To evaluate the integral $I = \int x \log x \, dx$, we utilize the method of Integration by Parts, which is defined by the formula:

$\int u(x)v'(x) \, dx = u(x)v(x) - \int u'(x)v(x) \, dx$ [Per the Integration by Parts Rule]

We apply the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) to select our functions:

• Let $u = \log x$ (Logarithmic function)
• Let $dv = x \, dx$ (Algebraic function)

Step 2: Differentiation and Integration of Selected Components

Differentiating $u$ with respect to $x$ and integrating $dv$:

• $du = \frac{d}{dx}(\log x) \, dx = \frac{1}{x} \, dx$
• $v = \int x \, dx = \frac{x^2}{2}$ [By the Power Rule of Integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1}$]

Step 3: Substitution into the Formula

Substituting these values into the Integration by Parts formula:

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

Step 4: Algebraic Simplification and Final Integration

Simplify the integrand within the integral expression:

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

Perform the final integration of $x$:

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

$I = \frac{x^2}{2} \log x - \frac{x^2}{4} + C$

Visual Representation of the Resulting Function

Step 5: Final Result

Factoring the expression for clarity:

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

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