Calculus Examples

$\int_{1}^{3} \frac{{(1 + \ln (x))}^{3}}{x} d x$

Step 1

Let $u = 1 + \ln (x)$ . Then $d u = \frac{1}{x} d x$ , so $x d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = 1 + \ln (x)$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $1 + \ln (x)$ .

$\frac{d}{d x} [1 + \ln (x)]$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $1 + \ln (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\ln (x)]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [\ln (x)]$

Step 1.1.2.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [\ln (x)]$

Step 1.1.3

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$0 + \frac{1}{x}$

Step 1.1.4

Add $0$ and $\frac{1}{x}$ .

$\frac{1}{x}$

Step 1.2

Substitute the lower limit in for $x$ in $u = 1 + \ln (x)$ .

$u_{lower} = 1 + \ln (1)$

Step 1.3

Simplify.

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Step 1.3.1

The natural logarithm of $1$ is $0$ .

$u_{lower} = 1 + 0$

Step 1.3.2

Add $1$ and $0$ .

$u_{lower} = 1$

Step 1.4

Substitute the upper limit in for $x$ in $u = 1 + \ln (x)$ .

$u_{upper} = 1 + \ln (3)$

Step 1.5

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 1$

$u_{upper} = 1 + \ln (3)$

Step 1.6

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$\int_{1}^{1 + \ln (3)} u^{3} d u$

Step 2

By the Power Rule, the integral of $u^{3}$ with respect to $u$ is $\frac{1}{4} u^{4}$ .

$\frac{1}{4} u^{4}]_{1}^{1 + \ln (3)}$

Step 3

Substitute and simplify.

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Step 3.1

Evaluate $\frac{1}{4} u^{4}$ at $1 + \ln (3)$ and at $1$ .

$(\frac{1}{4} {(1 + \ln (3))}^{4}) - \frac{1}{4} \cdot 1^{4}$

Step 3.2

Simplify.

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Step 3.2.1

Combine $\frac{1}{4}$ and ${(1 + \ln (3))}^{4}$ .

$\frac{{(1 + \ln (3))}^{4}}{4} - \frac{1}{4} \cdot 1^{4}$

Step 3.2.2

One to any power is one.

$\frac{{(1 + \ln (3))}^{4}}{4} - \frac{1}{4} \cdot 1$

Step 3.2.3

Multiply $- 1$ by $1$ .

$\frac{{(1 + \ln (3))}^{4}}{4} - \frac{1}{4}$

Step 3.2.4

Combine the numerators over the common denominator.

$\frac{{(1 + \ln (3))}^{4} - 1}{4}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$\frac{{(1 + \ln (3))}^{4} - 1}{4}$

Decimal Form:

$4.59918613 \dots$