Calculus Examples

$\int (9 - 2017 x^{2} \cdot 16 + 6 \sqrt[5]{x} + 12 e^{4 x} - \frac{5}{x}) d x$

Step 1

Remove parentheses.

$\int 9 - 2017 x^{2} \cdot 16 + 6 \sqrt[5]{x} + 12 e^{4 x} - \frac{5}{x} d x$

Step 2

Multiply $16$ by $- 2017$ .

$\int 9 - 32272 x^{2} + 6 \sqrt[5]{x} + 12 e^{4 x} - \frac{5}{x} d x$

Step 3

Split the single integral into multiple integrals.

$\int 9 d x + \int - 32272 x^{2} d x + \int 6 \sqrt[5]{x} d x + \int 12 e^{4 x} d x + \int - \frac{5}{x} d x$

Step 4

Apply the constant rule.

$9 x + C + \int - 32272 x^{2} d x + \int 6 \sqrt[5]{x} d x + \int 12 e^{4 x} d x + \int - \frac{5}{x} d x$

Step 5

Since $- 32272$ is constant with respect to $x$ , move $- 32272$ out of the integral.

$9 x + C - 32272 \int x^{2} d x + \int 6 \sqrt[5]{x} d x + \int 12 e^{4 x} d x + \int - \frac{5}{x} d x$

Step 6

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

$9 x + C - 32272 (\frac{1}{3} x^{3} + C) + \int 6 \sqrt[5]{x} d x + \int 12 e^{4 x} d x + \int - \frac{5}{x} d x$

Step 7

Since $6$ is constant with respect to $x$ , move $6$ out of the integral.

$9 x + C - 32272 (\frac{1}{3} x^{3} + C) + 6 \int \sqrt[5]{x} d x + \int 12 e^{4 x} d x + \int - \frac{5}{x} d x$

Step 8

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{x}$ as $x^{\frac{1}{5}}$ .

$9 x + C - 32272 (\frac{1}{3} x^{3} + C) + 6 \int x^{\frac{1}{5}} d x + \int 12 e^{4 x} d x + \int - \frac{5}{x} d x$

Step 9

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

$9 x + C - 32272 (\frac{1}{3} x^{3} + C) + 6 (\frac{5}{6} x^{\frac{6}{5}} + C) + \int 12 e^{4 x} d x + \int - \frac{5}{x} d x$

Step 10

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

$9 x + C - 32272 (\frac{1}{3} x^{3} + C) + 6 (\frac{5}{6} x^{\frac{6}{5}} + C) + 12 \int e^{4 x} d x + \int - \frac{5}{x} d x$

Step 11

Let $u = 4 x$ . Then $d u = 4 d x$ , so $\frac{1}{4} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 11.1

Let $u = 4 x$ . Find $\frac{d u}{d x}$ .

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Step 11.1.1

Differentiate $4 x$ .

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

Step 11.1.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

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

Step 11.1.3

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

$4 \cdot 1$

Step 11.1.4

Multiply $4$ by $1$ .

$4$

Step 11.2

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

$9 x + C - 32272 (\frac{1}{3} x^{3} + C) + 6 (\frac{5}{6} x^{\frac{6}{5}} + C) + 12 \int e^{u} \frac{1}{4} d u + \int - \frac{5}{x} d x$

Step 12

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

$9 x + C - 32272 (\frac{1}{3} x^{3} + C) + 6 (\frac{5}{6} x^{\frac{6}{5}} + C) + 12 (\frac{1}{4} \int e^{u} d u) + \int - \frac{5}{x} d x$

Step 13

Simplify.

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Step 13.1

Combine $\frac{1}{3}$ and $x^{3}$ .

$9 x + C - 32272 (\frac{x^{3}}{3} + C) + 6 (\frac{5}{6} x^{\frac{6}{5}} + C) + 12 (\frac{1}{4} \int e^{u} d u) + \int - \frac{5}{x} d x$

Step 13.2

Combine $\frac{5}{6}$ and $x^{\frac{6}{5}}$ .

$9 x + C - 32272 (\frac{x^{3}}{3} + C) + 6 (\frac{5 x^{\frac{6}{5}}}{6} + C) + 12 (\frac{1}{4} \int e^{u} d u) + \int - \frac{5}{x} d x$

Step 14

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

$9 x + C - 32272 (\frac{x^{3}}{3} + C) + 6 (\frac{5 x^{\frac{6}{5}}}{6} + C) + 12 (\frac{1}{4} (e^{u} + C)) + \int - \frac{5}{x} d x$

Step 15

Simplify.

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Step 15.1

Combine $\frac{1}{4}$ and $12$ .

$9 x + C - 32272 (\frac{x^{3}}{3} + C) + 6 (\frac{5 x^{\frac{6}{5}}}{6} + C) + \frac{12}{4} (e^{u} + C) + \int - \frac{5}{x} d x$

Step 15.2

Cancel the common factor of $12$ and $4$ .

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Step 15.2.1

Factor $4$ out of $12$ .

$9 x + C - 32272 (\frac{x^{3}}{3} + C) + 6 (\frac{5 x^{\frac{6}{5}}}{6} + C) + \frac{4 \cdot 3}{4} (e^{u} + C) + \int - \frac{5}{x} d x$

Step 15.2.2

Cancel the common factors.

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Step 15.2.2.1

Factor $4$ out of $4$ .

$9 x + C - 32272 (\frac{x^{3}}{3} + C) + 6 (\frac{5 x^{\frac{6}{5}}}{6} + C) + \frac{4 \cdot 3}{4 (1)} (e^{u} + C) + \int - \frac{5}{x} d x$

Step 15.2.2.2

Cancel the common factor.

$9 x + C - 32272 (\frac{x^{3}}{3} + C) + 6 (\frac{5 x^{\frac{6}{5}}}{6} + C) + \frac{4 \cdot 3}{4 \cdot 1} (e^{u} + C) + \int - \frac{5}{x} d x$

Step 15.2.2.3

Rewrite the expression.

$9 x + C - 32272 (\frac{x^{3}}{3} + C) + 6 (\frac{5 x^{\frac{6}{5}}}{6} + C) + \frac{3}{1} (e^{u} + C) + \int - \frac{5}{x} d x$

Step 15.2.2.4

Divide $3$ by $1$ .

$9 x + C - 32272 (\frac{x^{3}}{3} + C) + 6 (\frac{5 x^{\frac{6}{5}}}{6} + C) + 3 (e^{u} + C) + \int - \frac{5}{x} d x$

Step 16

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$9 x + C - 32272 (\frac{x^{3}}{3} + C) + 6 (\frac{5 x^{\frac{6}{5}}}{6} + C) + 3 (e^{u} + C) - \int \frac{5}{x} d x$

Step 17

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$9 x + C - 32272 (\frac{x^{3}}{3} + C) + 6 (\frac{5 x^{\frac{6}{5}}}{6} + C) + 3 (e^{u} + C) - (5 \int \frac{1}{x} d x)$

Step 18

Multiply $5$ by $- 1$ .

$9 x + C - 32272 (\frac{x^{3}}{3} + C) + 6 (\frac{5 x^{\frac{6}{5}}}{6} + C) + 3 (e^{u} + C) - 5 \int \frac{1}{x} d x$

Step 19

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

$9 x + C - 32272 (\frac{x^{3}}{3} + C) + 6 (\frac{5 x^{\frac{6}{5}}}{6} + C) + 3 (e^{u} + C) - 5 (\ln (| x |) + C)$

Step 20

Simplify.

$9 x - \frac{32272 x^{3}}{3} + 5 x^{\frac{6}{5}} + 3 e^{u} - 5 \ln (| x |) + C$

Step 21

Replace all occurrences of $u$ with $4 x$ .

$9 x - \frac{32272 x^{3}}{3} + 5 x^{\frac{6}{5}} + 3 e^{4 x} - 5 \ln (| x |) + C$

Step 22

Reorder terms.

$9 x - \frac{32272}{3} x^{3} + 5 x^{\frac{6}{5}} + 3 e^{4 x} - 5 \ln (| x |) + C$