Calculus Examples

$\int \frac{x^{3} \ln ({(2 + x^{4})}^{2})}{2 + x^{4}} d x$

Step 1

Rewrite $\frac{x^{3} \ln ({(2 + x^{4})}^{2})}{2 + x^{4}}$ as $\frac{x^{3} (2 \ln (2 + x^{4}))}{2 + x^{4}}$ .

$\int \frac{x^{3} (2 \ln (2 + x^{4}))}{2 + x^{4}} d x$

Step 2

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$2 \int \frac{x^{3} (\ln (2 + x^{4}))}{2 + x^{4}} d x$

Step 3

Let $u_{1} = 2 + x^{4}$ . Then $d u_{1} = 4 x^{3} d x$ , so $\frac{1}{4} d u_{1} = x^{3} d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 3.1

Let $u_{1} = 2 + x^{4}$ . Find $\frac{d u_{1}}{d x}$ .

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Step 3.1.1

Differentiate $2 + x^{4}$ .

$\frac{d}{d x} [2 + x^{4}]$

Step 3.1.2

By the Sum Rule, the derivative of $2 + x^{4}$ with respect to $x$ is $\frac{d}{d x} [2] + \frac{d}{d x} [x^{4}]$ .

$\frac{d}{d x} [2] + \frac{d}{d x} [x^{4}]$

Step 3.1.3

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [x^{4}]$

Step 3.1.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$0 + 4 x^{3}$

Step 3.1.5

Add $0$ and $4 x^{3}$ .

$4 x^{3}$

Step 3.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$2 \int \frac{\ln (u_{1})}{u_{1}} \cdot \frac{1}{4} d u_{1}$

Step 4

Simplify.

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Step 4.1

Multiply $\frac{\ln (u_{1})}{u_{1}}$ by $\frac{1}{4}$ .

$2 \int \frac{\ln (u_{1})}{u_{1} \cdot 4} d u_{1}$

Step 4.2

Move $4$ to the left of $u_{1}$ .

$2 \int \frac{\ln (u_{1})}{4 u_{1}} d u_{1}$

Step 5

Since $\frac{1}{4}$ is constant with respect to $u_{1}$ , move $\frac{1}{4}$ out of the integral.

$2 (\frac{1}{4} \int \frac{\ln (u_{1})}{u_{1}} d u_{1})$

Step 6

Simplify.

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Step 6.1

Combine $\frac{1}{4}$ and $2$ .

$\frac{2}{4} \int \frac{\ln (u_{1})}{u_{1}} d u_{1}$

Step 6.2

Cancel the common factor of $2$ and $4$ .

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Step 6.2.1

Factor $2$ out of $2$ .

$\frac{2 (1)}{4} \int \frac{\ln (u_{1})}{u_{1}} d u_{1}$

Step 6.2.2

Cancel the common factors.

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Step 6.2.2.1

Factor $2$ out of $4$ .

$\frac{2 \cdot 1}{2 \cdot 2} \int \frac{\ln (u_{1})}{u_{1}} d u_{1}$

Step 6.2.2.2

Cancel the common factor.

$\frac{2 \cdot 1}{2 \cdot 2} \int \frac{\ln (u_{1})}{u_{1}} d u_{1}$

Step 6.2.2.3

Rewrite the expression.

$\frac{1}{2} \int \frac{\ln (u_{1})}{u_{1}} d u_{1}$

Step 7

Let $u_{2} = \ln (u_{1})$ . Then $d u_{2} = \frac{1}{u_{1}} d u_{1}$ , so $u_{1} d u_{2} = d u_{1}$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 7.1

Let $u_{2} = \ln (u_{1})$ . Find $\frac{d u_{2}}{d u_{1}}$ .

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Step 7.1.1

Differentiate $\ln (u_{1})$ .

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

Step 7.1.2

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

$\frac{1}{u_{1}}$

Step 7.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\frac{1}{2} \int u_{2} d u_{2}$

Step 8

By the Power Rule, the integral of $u_{2}$ with respect to $u_{2}$ is $\frac{1}{2} {u_{2}}^{2}$ .

$\frac{1}{2} (\frac{1}{2} {u_{2}}^{2} + C)$

Step 9

Simplify.

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Step 9.1

Rewrite $\frac{1}{2} (\frac{1}{2} {u_{2}}^{2} + C)$ as $\frac{1}{2} \cdot \frac{1}{2} {u_{2}}^{2} + C$ .

$\frac{1}{2} \cdot \frac{1}{2} {u_{2}}^{2} + C$

Step 9.2

Simplify.

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Step 9.2.1

Multiply $\frac{1}{2}$ by $\frac{1}{2}$ .

$\frac{1}{2 \cdot 2} {u_{2}}^{2} + C$

Step 9.2.2

Multiply $2$ by $2$ .

$\frac{1}{4} {u_{2}}^{2} + C$

Step 10

Substitute back in for each integration substitution variable.

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Step 10.1

Replace all occurrences of $u_{2}$ with $\ln (u_{1})$ .

$\frac{1}{4} \ln^{2} (u_{1}) + C$

Step 10.2

Replace all occurrences of $u_{1}$ with $2 + x^{4}$ .

$\frac{1}{4} \ln^{2} (2 + x^{4}) + C$