Calculus Examples

$\int \frac{2 x^{2} e^{5 x^{3} + 1}}{3 - e^{5 x^{3} + 1}} d x$

Step 1

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$2 \int \frac{x^{2} e^{5 x^{3} + 1}}{3 - e^{5 x^{3} + 1}} d x$

Step 2

Let $u_{1} = 5 x^{3} + 1$ . Then $d u_{1} = 15 x^{2} d x$ , so $\frac{1}{15} d u_{1} = x^{2} d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.1

Let $u_{1} = 5 x^{3} + 1$ . Find $\frac{d u_{1}}{d x}$ .

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Step 2.1.1

Differentiate $5 x^{3} + 1$ .

$\frac{d}{d x} [5 x^{3} + 1]$

Step 2.1.2

By the Sum Rule, the derivative of $5 x^{3} + 1$ with respect to $x$ is $\frac{d}{d x} [5 x^{3}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [5 x^{3}] + \frac{d}{d x} [1]$

Step 2.1.3

Evaluate $\frac{d}{d x} [5 x^{3}]$ .

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Step 2.1.3.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{3}$ with respect to $x$ is $5 \frac{d}{d x} [x^{3}]$ .

$5 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [1]$

Step 2.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$5 (3 x^{2}) + \frac{d}{d x} [1]$

Step 2.1.3.3

Multiply $3$ by $5$ .

$15 x^{2} + \frac{d}{d x} [1]$

Step 2.1.4

Differentiate using the Constant Rule.

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Step 2.1.4.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$15 x^{2} + 0$

Step 2.1.4.2

Add $15 x^{2}$ and $0$ .

$15 x^{2}$

Step 2.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$2 \int \frac{e^{u_{1}}}{3 - e^{u_{1}}} \cdot \frac{1}{15} d u_{1}$

Step 3

Simplify.

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Step 3.1

Multiply $\frac{e^{u_{1}}}{3 - e^{u_{1}}}$ by $\frac{1}{15}$ .

$2 \int \frac{e^{u_{1}}}{(3 - e^{u_{1}}) \cdot 15} d u_{1}$

Step 3.2

Move $15$ to the left of $3 - e^{u_{1}}$ .

$2 \int \frac{e^{u_{1}}}{15 (3 - e^{u_{1}})} d u_{1}$

Step 4

Since $\frac{1}{15}$ is constant with respect to $u_{1}$ , move $\frac{1}{15}$ out of the integral.

$2 (\frac{1}{15} \int \frac{e^{u_{1}}}{3 - e^{u_{1}}} d u_{1})$

Step 5

Combine $\frac{1}{15}$ and $2$ .

$\frac{2}{15} \int \frac{e^{u_{1}}}{3 - e^{u_{1}}} d u_{1}$

Step 6

Let $u_{2} = 3 - e^{u_{1}}$ . Then $d u_{2} = - e^{u_{1}} d u_{1}$ , so $- \frac{1}{e^{u_{1}}} d u_{2} = d u_{1}$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 6.1

Let $u_{2} = 3 - e^{u_{1}}$ . Find $\frac{d u_{2}}{d u_{1}}$ .

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Step 6.1.1

Differentiate $3 - e^{u_{1}}$ .

$\frac{d}{d u_{1}} [3 - e^{u_{1}}]$

Step 6.1.2

Differentiate.

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Step 6.1.2.1

By the Sum Rule, the derivative of $3 - e^{u_{1}}$ with respect to $u_{1}$ is $\frac{d}{d u_{1}} [3] + \frac{d}{d u_{1}} [- e^{u_{1}}]$ .

$\frac{d}{d u_{1}} [3] + \frac{d}{d u_{1}} [- e^{u_{1}}]$

Step 6.1.2.2

Since $3$ is constant with respect to $u_{1}$ , the derivative of $3$ with respect to $u_{1}$ is $0$ .

$0 + \frac{d}{d u_{1}} [- e^{u_{1}}]$

Step 6.1.3

Evaluate $\frac{d}{d u_{1}} [- e^{u_{1}}]$ .

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Step 6.1.3.1

Since $- 1$ is constant with respect to $u_{1}$ , the derivative of $- e^{u_{1}}$ with respect to $u_{1}$ is $- \frac{d}{d u_{1}} [e^{u_{1}}]$ .

$0 - \frac{d}{d u_{1}} [e^{u_{1}}]$

Step 6.1.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$0 - e^{u_{1}}$

Step 6.1.4

Subtract $e^{u_{1}}$ from $0$ .

$- e^{u_{1}}$

Step 6.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\frac{2}{15} \int \frac{- 1}{u_{2}} d u_{2}$

Step 7

Move the negative in front of the fraction.

$\frac{2}{15} \int - \frac{1}{u_{2}} d u_{2}$

Step 8

Since $- 1$ is constant with respect to $u_{2}$ , move $- 1$ out of the integral.

$\frac{2}{15} (- \int \frac{1}{u_{2}} d u_{2})$

Step 9

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\frac{2}{15} (- (\ln (| u_{2} |) + C))$

Step 10

Simplify.

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Step 10.1

Simplify.

$\frac{2}{15} (- \ln (| u_{2} |)) + C$

Step 10.2

Combine $\frac{2}{15}$ and $\ln (| u_{2} |)$ .

$- \frac{2 \ln (| u_{2} |)}{15} + C$

Step 11

Substitute back in for each integration substitution variable.

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Step 11.1

Replace all occurrences of $u_{2}$ with $3 - e^{u_{1}}$ .

$- \frac{2 \ln (| 3 - e^{u_{1}} |)}{15} + C$

Step 11.2

Replace all occurrences of $u_{1}$ with $5 x^{3} + 1$ .

$- \frac{2 \ln (| 3 - e^{5 x^{3} + 1} |)}{15} + C$

Step 12

Reorder terms.

$- \frac{2}{15} \ln (| 3 - e^{5 x^{3} + 1} |) + C$