Calculus Examples

$f (x) = 5 e^{6 x} + \frac{- 3 x^{6} + 4 x}{x^{2}}$

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.

$F (x) = \int 5 e^{6 x} + \frac{- 3 x^{6} + 4 x}{x^{2}} d x$

Step 3

Simplify.

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Step 3.1

Factor $x$ out of $- 3 x^{6} + 4 x$ .

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Step 3.1.1

Factor $x$ out of $- 3 x^{6}$ .

$\int 5 e^{6 x} + \frac{x (- 3 x^{5}) + 4 x}{x^{2}} d x$

Step 3.1.2

Factor $x$ out of $4 x$ .

$\int 5 e^{6 x} + \frac{x (- 3 x^{5}) + x \cdot 4}{x^{2}} d x$

Step 3.1.3

Factor $x$ out of $x (- 3 x^{5}) + x \cdot 4$ .

$\int 5 e^{6 x} + \frac{x (- 3 x^{5} + 4)}{x^{2}} d x$

Step 3.2

Cancel the common factors.

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Step 3.2.1

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

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

Step 3.2.2

Cancel the common factor.

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

Step 3.2.3

Rewrite the expression.

$\int 5 e^{6 x} + \frac{- 3 x^{5} + 4}{x} d x$

Step 4

Split the single integral into multiple integrals.

$\int 5 e^{6 x} d x + \int \frac{- 3 x^{5} + 4}{x} d x$

Step 5

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

$5 \int e^{6 x} d x + \int \frac{- 3 x^{5} + 4}{x} d x$

Step 6

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 6.1

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

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Step 6.1.1

Differentiate $6 x$ .

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

Step 6.1.2

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 6.1.3

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

$6 \cdot 1$

Step 6.1.4

Multiply $6$ by $1$ .

$6$

Step 6.2

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

$5 \int e^{u} \frac{1}{6} d u + \int \frac{- 3 x^{5} + 4}{x} d x$

Step 7

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

$5 \int \frac{e^{u}}{6} d u + \int \frac{- 3 x^{5} + 4}{x} d x$

Step 8

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

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

Step 9

Combine $\frac{1}{6}$ and $5$ .

$\frac{5}{6} \int e^{u} d u + \int \frac{- 3 x^{5} + 4}{x} d x$

Step 10

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

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

Step 11

Divide $- 3 x^{5} + 4$ by $x$ .

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Step 11.1

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

$x$

$0$

$3 x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$4$

Step 11.2

Divide the highest order term in the dividend $- 3 x^{5}$ by the highest order term in divisor $x$ .

$3 x^{4}$

$x$

$0$

$3 x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$4$

Step 11.3

Multiply the new quotient term by the divisor.

$3 x^{4}$

$x$

$0$

$3 x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{5}$

$0$

Step 11.4

The expression needs to be subtracted from the dividend, so change all the signs in $- 3 x^{5} + 0$

$3 x^{4}$

$x$

$0$

$3 x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{5}$

$0$

Step 11.5

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

$3 x^{4}$

$x$

$0$

$3 x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{5}$

$0$

Step 11.6

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

$3 x^{4}$

$x$

$0$

$3 x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{5}$

$0$

$0 x^{3}$

$0 x^{2}$

Step 11.7

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

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

Step 12

Split the single integral into multiple integrals.

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

Step 13

Since $- 3$ is constant with respect to $x$ , move $- 3$ out of the integral.

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

Step 14

By the Power Rule, the integral of $x^{4}$ with respect to $x$ is $\frac{1}{5} x^{5}$ .

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

Step 15

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

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

Step 16

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

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

Step 17

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

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

Step 18

Simplify.

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

Step 19

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

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

Step 20

Reorder terms.

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

Step 21

The answer is the antiderivative of the function $f (x) = 5 e^{6 x} + \frac{- 3 x^{6} + 4 x}{x^{2}}$ .

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