Evaluate the following integrals:

Solution:

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

Expressing the integral $\int \frac{\mathrm{P}(\mathrm{x})}{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}} \mathrm{dx}=\int \mathrm{Q}(\mathrm{x}) \mathrm{dx}+\int \frac{\mathrm{R}(\mathrm{x})}{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}} \mathrm{dx}$

$\Rightarrow \int \frac{x^{2}\left(x^{4}+4\right)}{x^{2}+4} d x=\int\left(-\frac{80}{x^{2}+4}+x^{4}-4 x^{2}+20\right) d x$

$=-80 \int \frac{1}{x^{2}+4} d x+\int x^{4} d x-4 \int x^{2} d x+20 \int 1 d x$

Consider $\int \frac{1}{x^{2}+4} d x$

Let $u=1 / 2 x \rightarrow d x=2 d u$

$\Rightarrow \int \frac{1}{\mathrm{x}^{2}+4} \mathrm{dx}=\int \frac{2}{4 \mathrm{u}^{2}+4} \mathrm{du}$

$=\frac{1}{2} \int \frac{1}{\mathrm{u}^{2}+1} \mathrm{du}$

We know that $\int \frac{1}{x^{2}+1} d x=\tan ^{-1} x+c$'

$\Rightarrow \frac{1}{2} \int \frac{1}{\mathrm{u}^{2}+1} \mathrm{du}=\frac{\tan ^{-1} \mathrm{u}}{2}=\frac{\tan ^{-1}\left(\frac{\mathrm{x}}{2}\right)}{2}$

Then,

$\Rightarrow \int \frac{x^{2}\left(x^{4}+4\right)}{x^{2}+4} d x=-80 \int \frac{1}{x^{2}+4} d x+\int x^{4} d x-4 \int x^{2} d x+20 \int 1 d x$

We know that $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$ and $\int 1 d x=x+c$

$\Rightarrow-80\left(\frac{\tan ^{-1}\left(\frac{x}{2}\right)}{2}\right)+\frac{x^{5}}{5}-\frac{4 x^{3}}{3}+20 x+c$

$\Rightarrow-40 \tan ^{-1}\left(\frac{x}{2}\right)+\frac{x^{5}}{5}-\frac{4 x^{3}}{3}+20 x+c$

$\therefore \mathrm{I}=\int \frac{\mathrm{x}^{2}\left(\mathrm{x}^{4}+4\right)}{\mathrm{x}^{2}+4} \mathrm{dx}=-40 \tan ^{-1}\left(\frac{\mathrm{x}}{2}\right)+\frac{\mathrm{x}^{5}}{5}-\frac{4 \mathrm{x}^{3}}{3}+20 \mathrm{x}+\mathrm{c}$