# What is int 1 / (25 + x^2 ) dx?

Mar 5, 2016

$\frac{1}{5} T a {n}^{-} 1 \left(\frac{x}{5}\right) + c$

#### Explanation:

$\int \frac{1}{25 + {x}^{2}} \mathrm{dx}$

$= \int \frac{\mathrm{dx}}{{5}^{2} + {x}^{2}}$

$= I$

We use the rule:

$\int \frac{\mathrm{dx}}{{a}^{2} + {x}^{2}} = \frac{1}{a} T a {n}^{-} 1 \left(\frac{x}{a}\right) + c$

Here $a = 5$

$\implies I = \frac{1}{5} T a {n}^{-} 1 \left(\frac{x}{5}\right) + c$

Sep 16, 2016

$\frac{1}{5} \arctan \left(\frac{x}{5}\right) + C$

#### Explanation:

Alternatively, apply the substitution $x = 5 \tan \left(y\right)$. Note that this implies that $\mathrm{dx} = 5 {\sec}^{2} \left(y\right) \mathrm{dy}$.

$\int \frac{\mathrm{dx}}{25 + {x}^{2}} = \int \frac{5 {\sec}^{2} \left(y\right) \mathrm{dy}}{25 + 25 {\tan}^{2} \left(y\right)} = \frac{1}{5} \int \frac{{\sec}^{2} \left(y\right) \mathrm{dy}}{1 + {\tan}^{2} \left(y\right)}$

Using the identity $1 + {\tan}^{2} \left(y\right) = {\sec}^{2} \left(y\right)$:

$= \frac{1}{5} \int \frac{{\sec}^{2} \left(y\right) \mathrm{dy}}{{\sec}^{2} \left(y\right)} = \frac{1}{5} \int \mathrm{dy} = \frac{1}{5} y + C$

From $x = 5 \tan \left(y\right)$ we see that $y = \arctan \left(\frac{x}{5}\right)$:

$= \frac{1}{5} \arctan \left(\frac{x}{5}\right) + C$