# What is int (8x)/sqrt(16+x^2) dx?

##### 1 Answer
Jun 15, 2016

$8 \sqrt{16 + {x}^{2}} + C .$

#### Explanation:

Let $I = \int \frac{8 x}{\sqrt{16 + {x}^{2}}} \mathrm{dx} .$

We use Substitution $: 16 + {x}^{2} = {t}^{2.}$

$\therefore 2 x \mathrm{dx} = 2 t \mathrm{dt} .$, or, $x \mathrm{dx} = t \mathrm{dt} .$

Therefore, $I = \int \frac{8 x}{\sqrt{16 + {x}^{2}}} \mathrm{dx} , = \int \frac{8 t}{\sqrt{t}} ^ 2 \mathrm{dt} = 8 \int \mathrm{dt} = 8 t = 8 \sqrt{16 + {x}^{2}} + C .$

OR

We can use the Formula : $\int {\left(f \left(x\right)\right)}^{n} \cdot f ' \left(x\right) \mathrm{dx} = {\left(f \left(x\right)\right)}^{n + 1} / \left(n + 1\right) , n \ne - 1.$

Take $f \left(x\right) = \left(16 + {x}^{2}\right) ,$ $n = - \frac{1}{2} \ne - 1.$ Then, $f ' \left(x\right) = 2 x .$ Thus,
$I = \int \frac{8 x}{\sqrt{16 + {x}^{2}}} \mathrm{dx} = 4 \int {\left(16 + {x}^{2}\right)}^{- \frac{1}{2}} 2 x \mathrm{dx} = 4 \frac{{\left(16 + {x}^{2}\right)}^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} = 4 \frac{{\left(16 + {x}^{2}\right)}^{\frac{1}{2}}}{\frac{1}{2}} = 8 \sqrt{16 + {x}^{2}} + C ,$ as before!