Calculus Examples

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

Step 1

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

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

Step 2

Simplify.

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Step 2.1

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

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

Step 2.2

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

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

Step 3

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

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

Step 4

Multiply $4$ by $- 1$ .

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

Step 5

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

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Step 5.1

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

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Step 5.1.1

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

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

Step 5.1.2

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

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

Step 5.1.3

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

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Step 5.1.3.1

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

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

Step 5.1.3.2

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

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

Step 5.1.3.3

Multiply $2$ by $16$ .

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

Step 5.1.4

Differentiate using the Constant Rule.

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Step 5.1.4.1

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

$32 t + 0$

Step 5.1.4.2

Add $32 t$ and $0$ .

$32 t$

Step 5.2

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

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

Step 6

Simplify.

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Step 6.1

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

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

Step 6.2

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

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

Step 7

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

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

Step 8

Simplify.

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Step 8.1

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

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

Step 8.2

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

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Step 8.2.1

Factor $4$ out of $- 4$ .

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

Step 8.2.2

Cancel the common factors.

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Step 8.2.2.1

Factor $4$ out of $32$ .

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

Step 8.2.2.2

Cancel the common factor.

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

Step 8.2.2.3

Rewrite the expression.

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

Step 8.3

Move the negative in front of the fraction.

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

Step 9

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

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

Step 10

Simplify.

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

Step 11

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

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