Question

Integrate the following functions $w.r.t.x:$
$\frac{x}{{4x}^4-{2x}^2-3}$

Answer 1

We have,

$I=\int \frac{x}{{4x}^4-{2x}^2-3}dx$

Let

$t=x^2$

$\frac{dt}{dx}=2x$

$\frac{dt}{2}=xdx$

Therefore,

$I=\frac{1}{2}\int \frac{dt}{4t^2-2t-3}$

$I=\frac{1}{8}\int \frac{dt}{t^2-\frac{t}{2}-\frac{3}{4}}$

$I=\frac{1}{8}\int \frac{dt}{t^2-\frac{t}{2}+\frac{1}{16}-\frac{1}{16}-\frac{3}{4}}$

$I=\frac{1}{8}\int \frac{dt}{{(t-\frac{1}{4})}^2-{\left(\frac{\sqrt{13}}{4}\right)}^2}$

$I=\frac{1}{8}\left[\frac{1}{2\frac{\sqrt{13}}{4}}\ln \left(\frac{t-\frac{1}{4}-\frac{\sqrt{13}}{4}}{t-\frac{1}{4}+\frac{\sqrt{13}}{4}}\right)\right]+C$

$I=\frac{1}{8}\left[\frac{1}{\frac{\sqrt{13}}{2}}\ln \left(\frac{4t-1-\sqrt{13}}{4t-1+\sqrt{13}}\right)\right]+C$

$I=\frac{1}{4\sqrt{13}}\ln \left(\frac{4t-1-\sqrt{13}}{4t-1+\sqrt{13}}\right)+C$

On putting the value of $t$, we get

$I=\frac{1}{4\sqrt{13}}\ln \left(\frac{4x^2-1-\sqrt{13}}{4x^2-1+\sqrt{13}}\right)+C$

Hence, this is the answer.