# How do you integrate 1/(x^2 + 9)?

Jun 24, 2016

$\frac{1}{3} \arctan \left(\frac{x}{3}\right) + C$

#### Explanation:

We will try to put this in the form of the arctangent integral:

$\int \frac{1}{{u}^{2} + 1} \mathrm{du} = \arctan \left(u\right) + C$

So here, we see that:

$\int \frac{1}{{x}^{2} + 9} \mathrm{dx} = \int \frac{1}{9 \left({x}^{2} / 9 + 1\right)} \mathrm{dx} = \frac{1}{9} \int \frac{1}{{\left(\frac{x}{3}\right)}^{2} + 1} \mathrm{dx}$

Let $u = \frac{x}{3}$, implying that $\mathrm{du} = \frac{1}{3} \mathrm{dx}$:

$= \frac{1}{3} \int \frac{\frac{1}{3}}{{\left(\frac{x}{3}\right)}^{2} + 1} \mathrm{dx} = \frac{1}{3} \int \frac{1}{{u}^{2} + 1} \mathrm{du} = \frac{1}{3} \arctan \left(\frac{x}{3}\right) + C$

Jun 20, 2017

$\int \frac{\mathrm{dx}}{{x}^{2} + 9}$

Let $x = 3 \tan \theta$, implying that $\mathrm{dx} = 3 {\sec}^{2} \theta d \theta$.

$= \int \frac{3 {\sec}^{2} \theta d \theta}{9 {\tan}^{2} \theta + 9}$

Since ${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$:

$= \frac{1}{3} \int {\sec}^{2} \frac{\theta}{\sec} ^ 2 \theta d \theta = \frac{1}{3} \int d \theta = \frac{1}{3} \theta + C$

Reversing the original substitution $x = 3 \tan \theta$:

$= \frac{1}{3} {\tan}^{-} 1 \left(\frac{x}{3}\right) + C$