# What is int 2/(4+x^(2)) dx?

Apr 11, 2018

The answer is $= \arctan \left(\frac{x}{2}\right) + C$

#### Explanation:

Perform this integral by substitution

$4 + {x}^{2} = 4 \left(1 + {\left(\frac{x}{2}\right)}^{2}\right)$

Let $\tan u = \frac{x}{2}$

${\sec}^{2} u \mathrm{du} = \frac{1}{2} \mathrm{dx}$

$1 + {\tan}^{2} u = {\sec}^{2} u$

Therefore, the integral is

$\int \frac{2 \mathrm{dx}}{4 + {x}^{2}} = \int \frac{4 {\sec}^{2} u \mathrm{du}}{4 \left(1 + {\tan}^{2} u\right)}$

$= \int \frac{{\sec}^{2} u \mathrm{du}}{{\sec}^{2} u}$

$= \int \left(\mathrm{du}\right)$

$= u$

$= \arctan \left(\frac{x}{2}\right) + C$

Apr 11, 2018

The integral is equal to $\arctan \left(\frac{x}{2}\right) + C$.

#### Explanation:

To solve the integral, use the substitution $u = \frac{x}{2}$, which means $x = 2 u$ and $\mathrm{dx} = 2 \mathrm{du}$:

$\textcolor{w h i t e}{=} \int \frac{2}{4 + {x}^{2}}$ $\mathrm{dx}$

$= \int \frac{2}{4 + {\left(2 u\right)}^{2}}$ $2 \mathrm{du}$

$= \int \frac{4}{4 + 4 {u}^{2}}$ $\mathrm{du}$

$= \int \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} \left(1 + {u}^{2}\right)}$ $\mathrm{du}$

$= \int \frac{1}{1 + {u}^{2}}$ $\mathrm{du}$

$= \arctan \left(u\right) + C$

$= \arctan \left(\frac{x}{2}\right) + C$

That's the integral. Hope this helped!