Calculus Examples

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

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

Step 1

Remove parentheses.

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

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

Step 2

Multiply

16

$16$ by

- 2017

$- 2017$ .

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

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

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

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

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

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

Step 5

Since

- 32272

$- 32272$ is constant with respect to

x

$x$ , move

- 32272

$- 32272$ 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

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

Step 6

By the Power Rule, the integral of

x^{2}

$x^{2}$ with respect to

x

$x$ is

\frac{1}{3} x^{3}

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

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

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

Step 7

Since

6

$6$ is constant with respect to

x

$x$ , move

6

$6$ 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

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

Step 8

Use

\sqrt[n]{a^{x}} = a^{\frac{x}{n}}

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

\sqrt[5]{x}

$\sqrt[5]{x}$ as

x^{\frac{1}{5}}

$x^{\frac{1}{5}}$ .

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

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

Step 9

By the Power Rule, the integral of

x^{\frac{1}{5}}

$x^{\frac{1}{5}}$ with respect to

x

$x$ is

\frac{5}{6} x^{\frac{6}{5}}

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

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

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

Step 10

Since

12

$12$ is constant with respect to

x

$x$ , move

12

$12$ 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

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

Step 11

Let

u = 4 x

$u = 4 x$ . Then

d u = 4 d x

$d u = 4 d x$ , so

\frac{1}{4} d u = d x

$\frac{1}{4} d u = d x$ . Rewrite using

u

$u$ and

d

$d$

u

$u$ .

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Step 11.1

Let

u = 4 x

$u = 4 x$ . Find

\frac{d u}{d x}

$\frac{d u}{d x}$ .

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Step 11.1.1

Differentiate

4 x

$4 x$ .

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

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

Step 11.1.2

Since

4

$4$ is constant with respect to

x

$x$ , the derivative of

4 x

$4 x$ with respect to

x

$x$ is

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

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

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}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

4 \cdot 1

$4 \cdot 1$

Step 11.1.4

Multiply

4

$4$ by

1

$1$ .

4

$4$

4

$4$

Step 11.2

Rewrite the problem using

u

$u$ and

d u

$d u$ .

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

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

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

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

Step 12

Since

\frac{1}{4}

$\frac{1}{4}$ is constant with respect to

u

$u$ , move

\frac{1}{4}

$\frac{1}{4}$ 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

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

Step 13

Simplify.

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Step 13.1

Combine

\frac{1}{3}

$\frac{1}{3}$ and

x^{3}

$x^{3}$ .

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

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

Step 13.2

Combine

\frac{5}{6}

$\frac{5}{6}$ and

x^{\frac{6}{5}}

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

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

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

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

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

Step 14

The integral of

e^{u}

$e^{u}$ with respect to

u

$u$ is

e^{u}

$e^{u}$ .

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

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

Step 15

Simplify.

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Step 15.1

Combine

\frac{1}{4}

$\frac{1}{4}$ and

12

$12$ .

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

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

Step 15.2

Cancel the common factor of

12

$12$ and

4

$4$ .

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Step 15.2.1

Factor

4

$4$ out of

12

$12$ .

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

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

Step 15.2.2

Cancel the common factors.

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Step 15.2.2.1

Factor

4

$4$ out of

4

$4$ .

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

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

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.

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