# What is int x/ sqrt(x^2 - 8^2) dx?

Feb 4, 2016

$\int \frac{x}{\sqrt{{x}^{2} - 64}} \text{d} x = \sqrt{{x}^{2} - 64} + C$

where $C$ is a constant.

#### Explanation:

$\int \frac{x}{\sqrt{{x}^{2} - {8}^{2}}} \text{d"x = int x/(sqrt(x^2 - 64)) "d} x$

Let's use integration by substitution here.

Let $u = \sqrt{{x}^{2} - 64}$. We need to differentiate $u$ as next:

$\left(\text{d"u)/("d} x\right) = \frac{1}{2 \sqrt{{x}^{2} - 64}} \cdot 2 x = \frac{\cancel{2} x}{\cancel{2} \sqrt{{x}^{2} - 64}} = \frac{x}{\sqrt{{x}^{2} - 64}}$

Multiply by $\text{d} x$ to have just $\text{d} u$ on one side:

$\text{d"u = x / (sqrt(x^2 - 64)) "d} x$

As you can see, the substition seems to fit our integral perfectly. So we can start to substitute and solve the integral:

int x/(sqrt(x^2 - 64)) "d"x = int 1 * color(blue)(x/(sqrt(x^2 - 64)) "d"x)

... replace $\frac{x}{\sqrt{{x}^{2} - 64}} \text{d} x$ by $\text{d} u$ and all other occurences of $\sqrt{{x}^{2} - 64}$, if present, by $u$...

$= \int 1 \textcolor{w h i t e}{i i} \textcolor{b l u e}{\text{d} u}$

... the integral of $1$ is $u + C$ with a constant $C$...

$= u + C$

... re-substitute by replacing $u$ with $\sqrt{{x}^{2} - 64}$ ...

$= \sqrt{{x}^{2} - 64} + C$

Hope that this helped!