Question

Evaluate the following integrals:
$\int \frac{x^3-3x}{x^4+2x^2-4}dx$

Answer 1

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

$=\int \frac{x(x^2-3)}{x^4+2x^2-4}dx$

Let $x^2=t$, or, $2xdx=dt$

$$\begin{gathered} \Rightarrow I=\frac{1}{2}\int \frac{(t-3)}{t^2+2t-4}dt \\ =\frac{1}{4}\int \frac{2t-6}{t^2+2t-4}dt \\ =\frac{1}{4}\int \frac{2t+2-8}{t^2+2t-4}dt \\ =\frac{1}{4}\int \left(\frac{2t+2}{t^2+2t-4}-\frac{8}{t^2+2t-4}\right)dt \\ =\frac{1}{4}\left(\int \frac{2t+2}{t^2+2t-4}dt-\int \frac{8}{t^2+2t-4}dt\right) \end{gathered}$$

$\Rightarrow I=\frac{1}{4}\left(I_1+I_2\right)...\left(\mathrm{i}\right)$

Now,

$I_1=\int \frac{2t+2}{t^2+2t-4}dt$

Let $t^2+2t-4=u$

$$\begin{gathered} \mathrm{or},\left(2t+2\right)dt=du \\ \Rightarrow I_1=\int \frac{1}{u}du=\ln \left|u\right|+c_1 \\ \Rightarrow I_1=\ln \left|t^2+2t-4\right|+c_1 \\ \therefore I_1=\ln \left|x^4+2x^2-4\right|+c_1 \end{gathered}$$

Now,

$$\begin{gathered} I_2=\int \frac{-8}{(t+1{)}^2-5}dt \\ \Rightarrow I_2=\int \frac{8}{(\sqrt{5}{)}^2-(t+1{)}^2}dt \\ \therefore I_2=\frac{8}{2\sqrt{5}}\ln \left|\frac{\sqrt{5}+x^2+1}{\sqrt{5}-x^2-1}\right|+c_2 \end{gathered}$$

$$\begin{gathered} \mathrm{So},\mathrm{from}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get} \\ I=\frac{1}{4}\left[\ln \left|x^4+2x^2-4\right|+\frac{4}{\sqrt{5}}\ln \left|\frac{\sqrt{5}+x^2+1}{\sqrt{5}-x^2-1}\right|\right]+C \\ \therefore I=\frac{1}{4}\ln \left|x^4+2x^2-4\right|+\frac{1}{\sqrt{5}}\ln \left|\frac{\sqrt{5}+x^2+1}{\sqrt{5}-x^2-1}\right|+C \end{gathered}$$