Evaluate the integral:

Let, $I=\int \frac{x^{3}-3 x}{x^{4}+2 x^{2}-4} d x$

$I=\int \frac{\left(x^{2}-3\right) x}{\left(x^{2}\right)^{2}+2 x^{2}-4} d x$

If we assume $x^{2}$ to be an another variable, we can simplify the integral as derivative of $x^{2}$ i.e. $x$ is present in numerator.

Let, $x^{2}=u$

$\Rightarrow 2 \mathrm{x} \mathrm{dx}=\mathrm{du}$

$\Rightarrow \mathrm{x} \mathrm{dx}=1 / 2 \mathrm{du}$

$\therefore I=\frac{1}{2} \int \frac{\mathrm{u}-3}{\mathrm{u}^{2}+2 \mathrm{u}-4} \mathrm{du}$

As, $\frac{\mathrm{d}}{\mathrm{du}}\left(\mathrm{u}^{2}+2 \mathrm{u}-4\right)=2 \mathrm{u}+2$

$\therefore$ Let, $u-3=A(2 u+2)+B$

$\Rightarrow u-3=2 A u+2 A+B$

On comparing both sides -

We have,

$2 A=1 \Rightarrow A=1 / 2$

$2 A+B=-3 \Rightarrow B=-3-2 A=-4$

Hence,

$I=\int \frac{\frac{1}{2}(2 u+2)-4}{u^{2}+2 u-4} d u$

$\therefore I=\frac{1}{2} \int \frac{2 u+2}{u^{2}+2 u-4} d u-4 \int \frac{1}{u^{2}+2 u-4} d u$

Let, $I_{1}=\frac{1}{2} \int \frac{2 u+2}{u^{2}+2 u-4} d u$ and $I_{2}=\int \frac{1}{u^{2}+2 u-4} d u$

Now, $I=I_{1}-4 I_{2} \ldots .$ eqn 1

We will solve $I_{1}$ and $I_{2}$ individually.

$\mathrm{AS}, I_{1}=\frac{1}{2} \int \frac{2 \mathrm{u}+2}{\mathrm{u}^{2}+2 \mathrm{u}-4} \mathrm{du}$

Let $v=u^{2}+2 u-4 \Rightarrow d v=(2 u+2) d u$

$\therefore I_{1}$ reduces to $\frac{1}{2} \int \frac{\mathrm{dv}}{\mathrm{v}}$

Hence,

$\mathrm{I}_{1}=\frac{1}{2} \int \frac{\mathrm{dv}}{\mathrm{v}}=\log |\mathrm{u}|+\mathrm{C}\left\{\because \int \frac{\mathrm{dx}}{\mathrm{x}}=\log |\mathrm{x}|+\mathrm{C}\right\}$

On substituting value of $u$, we have:

$\mathrm{I}_{1}=\frac{1}{2} \log \left|\mathrm{u}^{2}+2 \mathrm{u}-4\right|+\mathrm{C} \ldots .$ eqn 2

As, $I_{2}=\int \frac{1}{u^{2}+2 u-4} d u$ and we don't have any derivative of function present in denominator.

$\therefore$ we will use some special integrals to solve the problem.

As denominator doesn't have any square root term. So one of the following two integrals will solve the problem.

i) $\int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C$ ii) $\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C$

Now we have to reduce $I_{2}$ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of $x$ is seen in denominator.

$\therefore I_{2}=\int \frac{1}{u^{2}+2 u-4} d u$

$\left.\Rightarrow\right|_{2}=\int \frac{1}{\left\{u^{2}+2(1) u+(1)^{2}\right\}-4-(1)^{2}} d u$

Using: $a^{2}+2 a b+b^{2}=(a+b)^{2}$

We have:

$I_{2}=\int \frac{1}{(u+1)^{2}-(\sqrt{5})^{2}} d u$

$I_{2}$ matches with $\int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C$

$\therefore \mathrm{I}_{2}=\frac{1}{2 \sqrt{5}} \log \left|\frac{\mathrm{u}+1-\sqrt{5}}{\mathrm{u}+1+\sqrt{5}}\right|+\mathrm{C} \ldots \mathrm{eqn} 3$

From eqn 1:

$I=I_{1}-4 I_{2}$

Using eqn 2 and eqn 3:

$I=\frac{1}{2} \log \left|u^{2}+2 u-4\right|-4\left(\frac{1}{2 \sqrt{5}} \log \left|\frac{u+1-\sqrt{5}}{u+1+\sqrt{5}}\right|\right)+C$

$I=\frac{1}{2} \log \left|u^{2}+2 u-4\right|-\frac{2}{\sqrt{5}} \log \left|\frac{u+1-\sqrt{5}}{u+1+\sqrt{5}}\right|+C$

Putting value of $u$ in 1 :

$I=\frac{1}{2} \log \left|x^{4}+2 x^{2}-4\right|-\frac{2}{\sqrt{5}} \log \left|\frac{x^{2}+1-\sqrt{5}}{x^{2}+1+\sqrt{5}}\right|+C$