Evaluate integral of arctan(4x) with respect to x

$\int \arctan (4 x) d x$

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \arctan (4 x)$ and $d v = 1$ .

$\arctan (4 x) x - \int x \frac{4}{16 x^{2} + 1} d x$

Simplify.

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Combine $x$ and $\frac{4}{16 x^{2} + 1}$ .

$\arctan (4 x) x - \int \frac{x \cdot 4}{16 x^{2} + 1} d x$

Move $4$ to the left of $x$ .

$\arctan (4 x) x - \int \frac{4 x}{16 x^{2} + 1} d x$

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

$\arctan (4 x) x - (4 \int \frac{x}{16 x^{2} + 1} d x)$

Multiply $4$ by $- 1$ .

$\arctan (4 x) x - 4 \int \frac{x}{16 x^{2} + 1} d x$

Let $u = 16 x^{2} + 1$ . Then $d u = 32 x d x$ , so $\frac{1}{32} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

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Let $u = 16 x^{2} + 1$ . Find $\frac{d u}{d x}$ .

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Differentiate $16 x^{2} + 1$ .

$\frac{d}{d x} [16 x^{2} + 1]$

By the Sum Rule, the derivative of $16 x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [16 x^{2}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [16 x^{2}] + \frac{d}{d x} [1]$

Evaluate $\frac{d}{d x} [16 x^{2}]$ .

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Since $16$ is constant with respect to $x$ , the derivative of $16 x^{2}$ with respect to $x$ is $16 \frac{d}{d x} [x^{2}]$ .

$16 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$

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

$16 (2 x) + \frac{d}{d x} [1]$

Multiply $2$ by $16$ .

$32 x + \frac{d}{d x} [1]$

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

$32 x + 0$

Add $32 x$ and $0$ .

$32 x$

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

$\arctan (4 x) x - 4 \int \frac{1}{u} \cdot \frac{1}{32} d u$

Simplify.

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Multiply $\frac{1}{u}$ and $\frac{1}{32}$ .

$\arctan (4 x) x - 4 \int \frac{1}{u \cdot 32} d u$

Move $32$ to the left of $u$ .

$\arctan (4 x) x - 4 \int \frac{1}{32 u} d u$

Since $\frac{1}{32}$ is constant with respect to $u$ , move $\frac{1}{32}$ out of the integral.

$\arctan (4 x) x - 4 (\frac{1}{32} \int \frac{1}{u} d u)$

Simplify.

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Combine $\frac{1}{32}$ and $- 4$ .

$\arctan (4 x) x + \frac{- 4}{32} \int \frac{1}{u} d u$

Cancel the common factor of $- 4$ and $32$ .

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Factor $4$ out of $- 4$ .

$\arctan (4 x) x + \frac{4 (- 1)}{32} \int \frac{1}{u} d u$

Cancel the common factors.

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Factor $4$ out of $32$ .

$\arctan (4 x) x + \frac{4 \cdot - 1}{4 \cdot 8} \int \frac{1}{u} d u$

Cancel the common factor.

$\arctan (4 x) x + \frac{4 \cdot - 1}{4 \cdot 8} \int \frac{1}{u} d u$

Rewrite the expression.

$\arctan (4 x) x + \frac{- 1}{8} \int \frac{1}{u} d u$

Move the negative in front of the fraction.

$\arctan (4 x) x - \frac{1}{8} \int \frac{1}{u} d u$

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

$\arctan (4 x) x - \frac{1}{8} (\ln (| u |) + C)$

Simplify.

$\arctan (4 x) x - \frac{1}{8} \ln (| u |) + C$

Replace all occurrences of $u$ with $16 x^{2} + 1$ .

$\arctan (4 x) x - \frac{1}{8} \ln (| 16 x^{2} + 1 |) + C$

Evaluate integral of arctan(4x) with respect to x

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