# Integral of (x+2)/((x^2-2)^0.5)

## \int\frac{x+2}{\sqrt{x^2-2}}dx

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$2\ln\left(\frac{\sqrt{x^2-2}}{\sqrt{2}}+\frac{x}{\sqrt{2}}\right)+\sqrt{x^2-2}+C_0$

## Step by step solution

Problem

$\int\frac{x+2}{\sqrt{x^2-2}}dx$
1

Split the fraction $\frac{x+2}{\sqrt{x^2-2}}$ in two terms with same denominator

$\int\left(\frac{2}{\sqrt{x^2-2}}+\frac{x}{\sqrt{x^2-2}}\right)dx$
2

The integral of a sum of two or more functions is equal to the sum of their integrals

$\int\frac{2}{\sqrt{x^2-2}}dx+\int\frac{x}{\sqrt{x^2-2}}dx$
3

Solve the integral $\int\frac{x}{\sqrt{x^2-2}}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=x^2-2 \\ du=2xdx\end{matrix}$
4

Isolate $dx$ in the previous equation

$\frac{du}{2x}=dx$
5

Substituting $u$ and $dx$ in the integral

$\int\frac{2}{\sqrt{x^2-2}}dx+\int\frac{1}{2\sqrt{u}}du$
6

Taking the constant out of the integral

$\int\frac{2}{\sqrt{x^2-2}}dx+\frac{1}{2}\int\frac{1}{\sqrt{u}}du$
7

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\int\frac{2}{\sqrt{x^2-2}}dx+\frac{1}{2}\int u^{-\frac{1}{2}}du$
8

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\int\frac{2}{\sqrt{x^2-2}}dx+\frac{1}{2}\cdot 2\sqrt{u}$
9

Substitute $u$ back for it's value, $x^2-2$

$\int\frac{2}{\sqrt{x^2-2}}dx+1\sqrt{x^2-2}$
10

Any expression multiplied by $1$ is equal to itself

$\int\frac{2}{\sqrt{x^2-2}}dx+\sqrt{x^2-2}$
11

Solve the integral $\int\frac{2}{\sqrt{x^2-2}}$ by trigonometric substitution using the substitution

$\begin{matrix}x=\sqrt{2}\sec\left(\theta\right) \\ dx=\sqrt{2}\tan\left(\theta\right)\sec\left(\theta\right)d\theta\end{matrix}$
12

Substituting in the original integral, we get

$\int\frac{\sqrt{8}\tan\left(\theta\right)\sec\left(\theta\right)}{\sqrt{2\sec\left(\theta\right)^2-2}}d\theta+\sqrt{x^2-2}$
13

Factor by the greatest common divisor $2$

$\int\frac{\sqrt{8}\tan\left(\theta\right)\sec\left(\theta\right)}{\sqrt{2\left(\sec\left(\theta\right)^2-1\right)}}d\theta+\sqrt{x^2-2}$
14

The power of a product is equal to the product of it's factors raised to the same power

$\int\frac{\sqrt{8}\tan\left(\theta\right)\sec\left(\theta\right)}{\sqrt{2}\sqrt{\sec\left(\theta\right)^2-1}}d\theta+\sqrt{x^2-2}$
15

Applying the trigonometric identity: $\tan\left(\theta\right)^2=\sec\left(\theta\right)^2-1$

$\int\frac{\sqrt{8}\tan\left(\theta\right)\sec\left(\theta\right)}{\sqrt{2}\tan\left(\theta\right)}d\theta+\sqrt{x^2-2}$
16

Simplifying the fraction by $\tan\left(\theta\right)$

$\int\frac{\sqrt{8}\sec\left(\theta\right)}{\sqrt{2}}d\theta+\sqrt{x^2-2}$
17

Taking the constant out of the integral

$\frac{\sqrt{1}}{2}\int\sqrt{8}\sec\left(\theta\right)d\theta+\sqrt{x^2-2}$
18

Taking the constant out of the integral

$\frac{\sqrt{1}}{2}\cdot \sqrt{8}\int\sec\left(\theta\right)d\theta+\sqrt{x^2-2}$
19

Multiply $\sqrt{8}$ times $\frac{\sqrt{1}}{2}$

$2\int\sec\left(\theta\right)d\theta+\sqrt{x^2-2}$
20

The integral of the secant function is given by the following formula, $\displaystyle\int\sec(x)dx=\ln\left|\sec(x)+\tan(x)\right|$

$2\ln\left(\tan\left(\theta\right)+\sec\left(\theta\right)\right)+\sqrt{x^2-2}$
21

Expressing the result of the integral in terms of the original variable

$2\ln\left(\frac{\sqrt{x^2-2}}{\sqrt{2}}+\frac{x}{\sqrt{2}}\right)+\sqrt{x^2-2}$
22

Add fraction's numerators with common denominators: $\frac{x}{\sqrt{2}}$ and $\frac{\sqrt{x^2-2}}{\sqrt{2}}$

$2\ln\left(\frac{\sqrt{x^2-2}+x}{\sqrt{2}}\right)+\sqrt{x^2-2}$
23

Split the fraction $\frac{x+\sqrt{x^2-2}}{\sqrt{2}}$ in two terms with same denominator

$2\ln\left(\frac{\sqrt{x^2-2}}{\sqrt{2}}+\frac{x}{\sqrt{2}}\right)+\sqrt{x^2-2}$
24

$2\ln\left(\frac{\sqrt{x^2-2}}{\sqrt{2}}+\frac{x}{\sqrt{2}}\right)+\sqrt{x^2-2}+C_0$

$2\ln\left(\frac{\sqrt{x^2-2}}{\sqrt{2}}+\frac{x}{\sqrt{2}}\right)+\sqrt{x^2-2}+C_0$

### Main topic:

Integration by trigonometric substitution

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