Calculus Examples

$f (x) = - 9 e^{- 9 x} + \frac{- 7 x + 5 x^{5}}{x^{2}}$

Step 1

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 2

Set up the integral to solve.

$F (x) = \int - 9 e^{- 9 x} + \frac{- 7 x + 5 x^{5}}{x^{2}} d x$

Step 3

Simplify.

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Step 3.1

Factor $x$ out of $- 7 x + 5 x^{5}$ .

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Step 3.1.1

Factor $x$ out of $- 7 x$ .

$\int - 9 e^{- 9 x} + \frac{x \cdot - 7 + 5 x^{5}}{x^{2}} d x$

Step 3.1.2

Factor $x$ out of $5 x^{5}$ .

$\int - 9 e^{- 9 x} + \frac{x \cdot - 7 + x (5 x^{4})}{x^{2}} d x$

Step 3.1.3

Factor $x$ out of $x \cdot - 7 + x (5 x^{4})$ .

$\int - 9 e^{- 9 x} + \frac{x (- 7 + 5 x^{4})}{x^{2}} d x$

Step 3.2

Cancel the common factors.

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Step 3.2.1

Factor $x$ out of $x^{2}$ .

$\int - 9 e^{- 9 x} + \frac{x (- 7 + 5 x^{4})}{x \cdot x} d x$

Step 3.2.2

Cancel the common factor.

$\int - 9 e^{- 9 x} + \frac{x (- 7 + 5 x^{4})}{x \cdot x} d x$

Step 3.2.3

Rewrite the expression.

$\int - 9 e^{- 9 x} + \frac{- 7 + 5 x^{4}}{x} d x$

Step 4

Split the single integral into multiple integrals.

$\int - 9 e^{- 9 x} d x + \int \frac{- 7 + 5 x^{4}}{x} d x$

Step 5

Since $- 9$ is constant with respect to $x$ , move $- 9$ out of the integral.

$- 9 \int e^{- 9 x} d x + \int \frac{- 7 + 5 x^{4}}{x} d x$

Step 6

Let $u = - 9 x$ . Then $d u = - 9 d x$ , so $- \frac{1}{9} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.1

Let $u = - 9 x$ . Find $\frac{d u}{d x}$ .

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Step 6.1.1

Differentiate $- 9 x$ .

$\frac{d}{d x} [- 9 x]$

Step 6.1.2

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x$ with respect to $x$ is $- 9 \frac{d}{d x} [x]$ .

$- 9 \frac{d}{d x} [x]$

Step 6.1.3

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

$- 9 \cdot 1$

Step 6.1.4

Multiply $- 9$ by $1$ .

$- 9$

Step 6.2

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

$- 9 \int e^{u} \frac{1}{- 9} d u + \int \frac{- 7 + 5 x^{4}}{x} d x$

Step 7

Simplify.

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Step 7.1

Move the negative in front of the fraction.

$- 9 \int e^{u} (- \frac{1}{9}) d u + \int \frac{- 7 + 5 x^{4}}{x} d x$

Step 7.2

Combine $e^{u}$ and $\frac{1}{9}$ .

$- 9 \int - \frac{e^{u}}{9} d u + \int \frac{- 7 + 5 x^{4}}{x} d x$

Step 8

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

$- 9 (- \int \frac{e^{u}}{9} d u) + \int \frac{- 7 + 5 x^{4}}{x} d x$

Step 9

Multiply $- 1$ by $- 9$ .

$9 \int \frac{e^{u}}{9} d u + \int \frac{- 7 + 5 x^{4}}{x} d x$

Step 10

Since $\frac{1}{9}$ is constant with respect to $u$ , move $\frac{1}{9}$ out of the integral.

$9 (\frac{1}{9} \int e^{u} d u) + \int \frac{- 7 + 5 x^{4}}{x} d x$

Step 11

Simplify.

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Step 11.1

Combine $\frac{1}{9}$ and $9$ .

$\frac{9}{9} \int e^{u} d u + \int \frac{- 7 + 5 x^{4}}{x} d x$

Step 11.2

Cancel the common factor of $9$ .

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Step 11.2.1

Cancel the common factor.

$\frac{9}{9} \int e^{u} d u + \int \frac{- 7 + 5 x^{4}}{x} d x$

Step 11.2.2

Rewrite the expression.

$1 \int e^{u} d u + \int \frac{- 7 + 5 x^{4}}{x} d x$

Step 11.3

Multiply $\int e^{u} d u$ by $1$ .

$\int e^{u} d u + \int \frac{- 7 + 5 x^{4}}{x} d x$

Step 12

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

$e^{u} + C + \int \frac{- 7 + 5 x^{4}}{x} d x$

Step 13

Reorder $- 7$ and $5 x^{4}$ .

$e^{u} + C + \int \frac{5 x^{4} - 7}{x} d x$

Step 14

Divide $5 x^{4} - 7$ by $x$ .

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Step 14.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$0$

$5 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$7$

Step 14.2

Divide the highest order term in the dividend $5 x^{4}$ by the highest order term in divisor $x$ .

$5 x^{3}$

$x$

$0$

$5 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$7$

Step 14.3

Multiply the new quotient term by the divisor.

$5 x^{3}$

$x$

$0$

$5 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$7$

$5 x^{4}$

$0$

Step 14.4

The expression needs to be subtracted from the dividend, so change all the signs in $5 x^{4} + 0$

$5 x^{3}$

$x$

$0$

$5 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$7$

$5 x^{4}$

$0$

Step 14.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$5 x^{3}$

$x$

$0$

$5 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$7$

$5 x^{4}$

$0$

Step 14.6

Pull the next term from the original dividend down into the current dividend.

$5 x^{3}$

$x$

$0$

$5 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$7$

$5 x^{4}$

$0$

$0 x^{2}$

$0 x$

Step 14.7

The final answer is the quotient plus the remainder over the divisor.

$e^{u} + C + \int 5 x^{3} - \frac{7}{x} d x$

Step 15

Split the single integral into multiple integrals.

$e^{u} + C + \int 5 x^{3} d x + \int - \frac{7}{x} d x$

Step 16

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

$e^{u} + C + 5 \int x^{3} d x + \int - \frac{7}{x} d x$

Step 17

By the Power Rule, the integral of $x^{3}$ with respect to $x$ is $\frac{1}{4} x^{4}$ .

$e^{u} + C + 5 (\frac{1}{4} x^{4} + C) + \int - \frac{7}{x} d x$

Step 18

Combine $\frac{1}{4}$ and $x^{4}$ .

$e^{u} + C + 5 (\frac{x^{4}}{4} + C) + \int - \frac{7}{x} d x$

Step 19

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

$e^{u} + C + 5 (\frac{x^{4}}{4} + C) - \int \frac{7}{x} d x$

Step 20

Since $7$ is constant with respect to $x$ , move $7$ out of the integral.

$e^{u} + C + 5 (\frac{x^{4}}{4} + C) - (7 \int \frac{1}{x} d x)$

Step 21

Multiply $7$ by $- 1$ .

$e^{u} + C + 5 (\frac{x^{4}}{4} + C) - 7 \int \frac{1}{x} d x$

Step 22

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

$e^{u} + C + 5 (\frac{x^{4}}{4} + C) - 7 (\ln (| x |) + C)$

Step 23

Simplify.

$e^{u} + \frac{5 x^{4}}{4} - 7 \ln (| x |) + C$

Step 24

Replace all occurrences of $u$ with $- 9 x$ .

$e^{- 9 x} + \frac{5 x^{4}}{4} - 7 \ln (| x |) + C$

Step 25

Reorder terms.

$e^{- 9 x} + \frac{5}{4} x^{4} - 7 \ln (| x |) + C$

Step 26

The answer is the antiderivative of the function $f (x) = - 9 e^{- 9 x} + \frac{- 7 x + 5 x^{5}}{x^{2}}$ .

$F (x) =$ $e^{- 9 x} + \frac{5}{4} x^{4} - 7 \ln (| x |) + C$