Calculus Examples

$\int 2^{4 x^{2}} x d x$

Step 1

Let $u = 4 x^{2}$ . Then $d u = 8 x d x$ , so $\frac{1}{8} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = 4 x^{2}$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $4 x^{2}$ .

$\frac{d}{d x} [4 x^{2}]$

Step 1.1.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2}]$ .

$4 \frac{d}{d x} [x^{2}]$

Step 1.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$4 (2 x)$

Step 1.1.4

Multiply $2$ by $4$ .

$8 x$

Step 1.2

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

$\int 2^{u} \frac{1}{8} d u$

Step 2

Combine $2^{u}$ and $\frac{1}{8}$ .

$\int \frac{2^{u}}{8} d u$

Step 3

Since $\frac{1}{8}$ is constant with respect to $u$ , move $\frac{1}{8}$ out of the integral.

$\frac{1}{8} \int 2^{u} d u$

Step 4

The integral of $2^{u}$ with respect to $u$ is $\frac{2^{u}}{\ln (2)}$ .

$\frac{1}{8} (\frac{2^{u}}{\ln (2)} + C)$

Step 5

Simplify.

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Step 5.1

Rewrite $\frac{1}{8} (\frac{2^{u}}{\ln (2)} + C)$ as $\frac{1}{8} \cdot \frac{1}{\ln (2)} \cdot 2^{u} + C$ .

$\frac{1}{8} \cdot \frac{1}{\ln (2)} \cdot 2^{u} + C$

Step 5.2

Multiply $\frac{1}{8}$ by $\frac{1}{\ln (2)}$ .

$\frac{1}{8 \ln (2)} \cdot 2^{u} + C$

Step 6

Replace all occurrences of $u$ with $4 x^{2}$ .

$\frac{1}{8 \ln (2)} \cdot 2^{4 x^{2}} + C$