Calculus Examples

$\int x^{2} {(2 x^{3} - 1)}^{4} d x$

Step 1

Let $u = 2 x^{3} - 1$ . Then $d u = 6 x^{2} d x$ , so $\frac{1}{6} d u = x^{2} d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = 2 x^{3} - 1$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $2 x^{3} - 1$ .

$\frac{d}{d x} [2 x^{3} - 1]$

Step 1.1.2

By the Sum Rule, the derivative of $2 x^{3} - 1$ with respect to $x$ is $\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [- 1]$ .

$\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [- 1]$

Step 1.1.3

Evaluate $\frac{d}{d x} [2 x^{3}]$ .

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Step 1.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x^{3}$ with respect to $x$ is $2 \frac{d}{d x} [x^{3}]$ .

$2 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 1]$

Step 1.1.3.2

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

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

Step 1.1.3.3

Multiply $3$ by $2$ .

$6 x^{2} + \frac{d}{d x} [- 1]$

Step 1.1.4

Differentiate using the Constant Rule.

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Step 1.1.4.1

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

$6 x^{2} + 0$

Step 1.1.4.2

Add $6 x^{2}$ and $0$ .

$6 x^{2}$

Step 1.2

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

$\int u^{4} \frac{1}{6} d u$

Step 2

Combine $u^{4}$ and $\frac{1}{6}$ .

$\int \frac{u^{4}}{6} d u$

Step 3

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

$\frac{1}{6} \int u^{4} d u$

Step 4

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

$\frac{1}{6} (\frac{1}{5} u^{5} + C)$

Step 5

Simplify.

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Step 5.1

Rewrite $\frac{1}{6} (\frac{1}{5} u^{5} + C)$ as $\frac{1}{6} \cdot \frac{1}{5} u^{5} + C$ .

$\frac{1}{6} \cdot \frac{1}{5} u^{5} + C$

Step 5.2

Rewrite $\frac{1}{6} \cdot \frac{1}{5} u^{5} + C$ as $\frac{1}{30} u^{5} + C$ .

$\frac{1}{30} u^{5} + C$

Step 6

Replace all occurrences of $u$ with $2 x^{3} - 1$ .

$\frac{1}{30} {(2 x^{3} - 1)}^{5} + C$