Question

Integrate the function: $\frac{3x}{1+2x^4}$

Answer 1

To find: $\int \frac{3x}{1+2x^4}.dx$

Let $x^2=t$
Differentiating both sides w.r.t $x$
$2x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{2}$

Now,
$\int \frac{3x}{1+2x^4}.dx=\frac{3}{2}\int \frac{1}{1+2t^2}.dx$

$\frac{3}{4}\int \frac{1}{\frac{1}{2}+t^2}.dx=\frac{3}{4}\int \frac{1}{{\left(\frac{1}{\sqrt{2}}\right)}^2+t^2}.dx$

$=\frac{3}{4}[\frac{1}{\frac{1}{\sqrt{2}}}.{\tan }^{-1}\left(\frac{t}{\frac{1}{\sqrt{2}}}\right)]+C$

$[\because \int \frac{1}{a^2+x^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C]$

$=\frac{3}{4}\left[\sqrt{2}{\tan }^{-1}\right(\sqrt{2}t\left)\right]+C$

$=\frac{3}{2\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}t\right)+C$

$=\frac{3}{2\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}{x}^2\right)+C$

where $C$ is a constant of integration.