Calculus Examples

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

Step 1

Let $u_{1} = x + 4$ . Then $d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 1.1

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

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Step 1.1.1

Differentiate $x + 4$ .

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

Step 1.1.2

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

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

Step 1.1.3

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

$1 + \frac{d}{d x} [4]$

Step 1.1.4

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$1 + 0$

Step 1.1.5

Add $1$ and $0$ .

$1$

Step 1.2

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

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

Step 2

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 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.2

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

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

Step 3

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

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

Step 4

Substitute back in for each integration substitution variable.

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Step 4.1

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

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

Step 4.2

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

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