Calculus Examples

$\int \frac{v + 1}{- 1 - v^{2}} d v$

Step 1

Split the fraction $\frac{v + 1}{- 1 - v^{2}}$ into two fractions.

$\int \frac{v}{- 1 - v^{2}} + \frac{1}{- 1 - v^{2}} d v$

Step 2

Split the single integral into multiple integrals.

$\int \frac{v}{- 1 - v^{2}} d v + \int \frac{1}{- 1 - v^{2}} d v$

Step 3

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

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Step 3.1

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

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Step 3.1.1

Differentiate $- 1 - v^{2}$ .

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

Step 3.1.2

Differentiate.

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Step 3.1.2.1

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

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

Step 3.1.2.2

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

$0 + \frac{d}{d v} [- v^{2}]$

Step 3.1.3

Evaluate $\frac{d}{d v} [- v^{2}]$ .

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Step 3.1.3.1

Since $- 1$ is constant with respect to $v$ , the derivative of $- v^{2}$ with respect to $v$ is $- \frac{d}{d v} [v^{2}]$ .

$0 - \frac{d}{d v} [v^{2}]$

Step 3.1.3.2

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

$0 - (2 v)$

Step 3.1.3.3

Multiply $2$ by $- 1$ .

$0 - 2 v$

Step 3.1.4

Subtract $2 v$ from $0$ .

$- 2 v$

Step 3.2

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

$\int \frac{1}{u} \cdot \frac{1}{- 2} d u + \int \frac{1}{- 1 - v^{2}} d v$

Step 4

Simplify.

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Step 4.1

Move the negative in front of the fraction.

$\int \frac{1}{u} (- \frac{1}{2}) d u + \int \frac{1}{- 1 - v^{2}} d v$

Step 4.2

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$\int - \frac{1}{u \cdot 2} d u + \int \frac{1}{- 1 - v^{2}} d v$

Step 4.3

Move $2$ to the left of $u$ .

$\int - \frac{1}{2 u} d u + \int \frac{1}{- 1 - v^{2}} d v$

Step 5

Since $- 1$ is constant with respect to $u$ , move $- 1$ out of the integral.

$- \int \frac{1}{2 u} d u + \int \frac{1}{- 1 - v^{2}} d v$

Step 6

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$- (\frac{1}{2} \int \frac{1}{u} d u) + \int \frac{1}{- 1 - v^{2}} d v$

Step 7

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

$- \frac{1}{2} (\ln (| u |) + C) + \int \frac{1}{- 1 - v^{2}} d v$

Step 8

Factor $- 1$ out of $- 1 - v^{2}$ .

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Step 8.1

Rewrite $- 1$ as $- 1 (1)$ .

$- \frac{1}{2} (\ln (| u |) + C) + \int \frac{1}{- 1 \cdot 1 - v^{2}} d v$

Step 8.2

Factor $- 1$ out of $- 1 \cdot 1$ .

$- \frac{1}{2} (\ln (| u |) + C) + \int \frac{1}{- 1 (1) - v^{2}} d v$

Step 8.3

Factor $- 1$ out of $- v^{2}$ .

$- \frac{1}{2} (\ln (| u |) + C) + \int \frac{1}{- 1 (1) - (v^{2})} d v$

Step 8.4

Factor $- 1$ out of $- 1 (1) - (v^{2})$ .

$- \frac{1}{2} (\ln (| u |) + C) + \int \frac{1}{- 1 (1 + v^{2})} d v$

Step 9

Simplify.

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Step 9.1

Rewrite $- 1 (1 + v^{2})$ as $- (1 + v^{2})$ .

$- \frac{1}{2} (\ln (| u |) + C) + \int \frac{1}{- (1 + v^{2})} d v$

Step 9.2

Cancel the common factor of $1$ and $- 1$ .

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Step 9.2.1

Rewrite $1$ as $- 1 (- 1)$ .

$- \frac{1}{2} (\ln (| u |) + C) + \int \frac{- 1 (- 1)}{- (1 + v^{2})} d v$

Step 9.2.2

Move the negative in front of the fraction.

$- \frac{1}{2} (\ln (| u |) + C) + \int - \frac{1}{1 + v^{2}} d v$

Step 10

Since $- 1$ is constant with respect to $v$ , move $- 1$ out of the integral.

$- \frac{1}{2} (\ln (| u |) + C) - \int \frac{1}{1 + v^{2}} d v$

Step 11

Rewrite $1$ as $1^{2}$ .

$- \frac{1}{2} (\ln (| u |) + C) - \int \frac{1}{1^{2} + v^{2}} d v$

Step 12

The integral of $\frac{1}{1^{2} + v^{2}}$ with respect to $v$ is $\arctan (v) + C$ .

$- \frac{1}{2} (\ln (| u |) + C) - (\arctan (v) + C)$

Step 13

Simplify.

$- \frac{1}{2} \ln (| u |) - \arctan (v) + C$

Step 14

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

$- \frac{1}{2} \ln (| - 1 - v^{2} |) - \arctan (v) + C$