Calculus Examples

$\int 12 x^{- 3} + e^{6 x} d x$

Step 1

Split the single integral into multiple integrals.

$\int 12 x^{- 3} d x + \int e^{6 x} d x$

Step 2

Since $12$ is constant with respect to $x$ , move $12$ out of the integral.

$12 \int x^{- 3} d x + \int e^{6 x} d x$

Step 3

By the Power Rule, the integral of $x^{- 3}$ with respect to $x$ is $- \frac{1}{2} x^{- 2}$ .

$12 (- \frac{1}{2} x^{- 2} + C) + \int e^{6 x} d x$

Step 4

Simplify.

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Step 4.1

Combine $x^{- 2}$ and $\frac{1}{2}$ .

$12 (- \frac{x^{- 2}}{2} + C) + \int e^{6 x} d x$

Step 4.2

Move $x^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$12 (- \frac{1}{2 x^{2}} + C) + \int e^{6 x} d x$

Step 5

Let $u = 6 x$ . Then $d u = 6 d x$ , so $\frac{1}{6} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1

Let $u = 6 x$ . Find $\frac{d u}{d x}$ .

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Step 5.1.1

Differentiate $6 x$ .

$\frac{d}{d x} [6 x]$

Step 5.1.2

Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$6 \frac{d}{d x} [x]$

Step 5.1.3

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

$6 \cdot 1$

Step 5.1.4

Multiply $6$ by $1$ .

$6$

Step 5.2

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

$12 (- \frac{1}{2 x^{2}} + C) + \int e^{u} \frac{1}{6} d u$

Step 6

Combine $e^{u}$ and $\frac{1}{6}$ .

$12 (- \frac{1}{2 x^{2}} + C) + \int \frac{e^{u}}{6} d u$

Step 7

Since $\frac{1}{6}$ is constant with respect to $u$ , move $\frac{1}{6}$ out of the integral.

$12 (- \frac{1}{2 x^{2}} + C) + \frac{1}{6} \int e^{u} d u$

Step 8

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$12 (- \frac{1}{2 x^{2}} + C) + \frac{1}{6} (e^{u} + C)$

Step 9

Simplify.

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Step 9.1

Simplify.

$\frac{- 6}{x^{2}} + \frac{1}{6} e^{u} + C$

Step 9.2

Move the negative in front of the fraction.

$- \frac{6}{x^{2}} + \frac{1}{6} e^{u} + C$

Step 10

Replace all occurrences of $u$ with $6 x$ .

$- \frac{6}{x^{2}} + \frac{1}{6} e^{6 x} + C$