Calculus Examples

$\int {(2 + e^{x})}^{- 6} e^{x} d x$

Step 1

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

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Step 1.1

Let $u = 2 + e^{x}$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $2 + e^{x}$ .

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

Step 1.1.2

Differentiate.

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Step 1.1.2.1

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

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

Step 1.1.2.2

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

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

Step 1.1.3

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

$0 + e^{x}$

Step 1.1.4

Add $0$ and $e^{x}$ .

$e^{x}$

Step 1.2

Rewrite the problem using $u$ and $d u$ .

$\int u^{- 6} d u$

Step 2

By the Power Rule, the integral of $u^{- 6}$ with respect to $u$ is $- \frac{1}{5} u^{- 5}$ .

$- \frac{1}{5} u^{- 5} + C$

Step 3

Simplify.

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Step 3.1

Rewrite $- \frac{1}{5} u^{- 5} + C$ as $- \frac{1}{5} \cdot \frac{1}{u^{5}} + C$ .

$- \frac{1}{5} \cdot \frac{1}{u^{5}} + C$

Step 3.2

Simplify.

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Step 3.2.1

Multiply $\frac{1}{u^{5}}$ by $\frac{1}{5}$ .

$- \frac{1}{u^{5} \cdot 5} + C$

Step 3.2.2

Move $5$ to the left of $u^{5}$ .

$- \frac{1}{5 u^{5}} + C$

Step 4

Replace all occurrences of $u$ with $2 + e^{x}$ .

$- \frac{1}{5 {(2 + e^{x})}^{5}} + C$