# Integrate the following int t/(t^4 +2) dt ?

Feb 19, 2017

$\frac{1}{2 \sqrt{2}} {\tan}^{-} 1 \left({t}^{2} / \sqrt{2}\right) + C$

#### Explanation:

Let ${t}^{2} = \sqrt{2} \tan \theta$. This implies that $2 t \mathrm{dt} = \sqrt{2} {\sec}^{2} \theta d \theta$. Then:

$I = \frac{1}{2} \int \frac{2 t \mathrm{dt}}{{\left({t}^{2}\right)}^{2} + 2} = \frac{1}{2} \int \frac{\sqrt{2} {\sec}^{2} \theta d \theta}{{\left(\sqrt{2} \tan \theta\right)}^{2} + 2}$

Continuing on and using ${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$:

$I = \frac{1}{\sqrt{2}} \int {\sec}^{2} \frac{\theta}{2 {\tan}^{2} \theta + 2} d \theta = \frac{1}{2 \sqrt{2}} \int {\sec}^{2} \frac{\theta}{\sec} ^ 2 \theta d \theta = \frac{1}{2 \sqrt{2}} \int d \theta$

We're working in terms of $\theta$:

$I = \frac{1}{2 \sqrt{2}} \theta + C$

From ${t}^{2} = \sqrt{2} \tan \theta$ we see that $\theta = {\tan}^{-} 1 \left({t}^{2} / \sqrt{2}\right)$:

$I = \frac{1}{2 \sqrt{2}} {\tan}^{-} 1 \left({t}^{2} / \sqrt{2}\right) + C$

I solved this way: