Evaluate the following integrals:

Solution:

Given $I=\int \frac{x^{2}}{x^{2}+7 x+10} d x$

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

$\Rightarrow \int \frac{x^{2}}{x^{2}+7 x+10} d x=\int\left(\frac{-7 x-10}{x^{2}+7 x+10}+1\right) d x$

$=-\int \frac{7 x+10}{x^{2}+7 x+10} d x+\int 1 d x$

Consider $\int \frac{7 x+10}{x^{2}+7 x+10} d x$

Let $7 x+10=\frac{7}{2}(2 x+7)-\frac{29}{2}$ and split,

$\Rightarrow \int \frac{7 x+10}{x^{2}+7 x+10} d x=\int\left(\frac{7(2 x+7)}{2\left(x^{2}+7 x+10\right)}-\frac{29}{2\left(x^{2}+7 x+10\right)}\right) d x$

$=\frac{7}{2} \int \frac{2 x+7}{x^{2}+7 x+10} d x-\frac{29}{2} \int \frac{1}{x^{2}+7 x+10} d x$

Consider $\int \frac{2 x+7}{x^{2}+7 x+10} d x$

Let $u=x^{2}+7 x+10 \rightarrow d x=\frac{1}{2 x+7} d u$

$\Rightarrow \int \frac{2 x+7}{\left(x^{2}+7 x+10\right)} d x=\int \frac{2 x+7}{u} \frac{1}{2 x+7} d u$

$=\int \frac{1}{u} d u$

We know that $\int \frac{1}{x} d x=\log |x|+c$

$\Rightarrow \int \frac{1}{u} d u=\log |u|=\log \left|x^{2}+7 x+10\right|$

Now consider $\int \frac{1}{x^{2}+7 x+10} d x$

$\Rightarrow \int \frac{1}{x^{2}+7 x+10} d x=\int \frac{1}{(x+2)(x+5)} d x$

By partial fraction decomposition,

$\Rightarrow \frac{1}{(x+2)(x+5)}=\frac{A}{x+2}+\frac{B}{x+5}$

$\Rightarrow 1=A(x+2)+B(x+5)$

$\Rightarrow 1=A x+2 A+B x+5 B$

$\Rightarrow 1=(A+B) x+(2 A+5 B)$

$\Rightarrow A+B=0$ and $2 A+5 B=1$

Solving the two equations,

$\Rightarrow 2 A+2 B=0$

$2 A+5 B=1$

$-3 B=-1$

$\therefore B=1 / 3$ and $A=-1 / 3$

$\Rightarrow \int \frac{1}{(x+2)(x+5)} d x=\int\left(\frac{-1}{3(x+2)}+\frac{1}{3(x+5)}\right) d x$

$=-\frac{1}{3} \int \frac{1}{x+2} d x+\frac{1}{3} \int \frac{1}{x+5} d x$

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

Let $u=x+2 \rightarrow d x=d u$

$\Rightarrow \int \frac{1}{x+2} d x=\int \frac{1}{u} d u$

We know that $\int \frac{1}{x} d x=\log |x|+c$

$\Rightarrow \int \frac{1}{\mathrm{u}} \mathrm{du}=\log |\mathrm{u}|=\log |\mathrm{x}+2|$

Similarly $\int \frac{1}{x+5} d x$

Let $u=x+5 \rightarrow d x=d u$

$\Rightarrow \int \frac{1}{x+5} d x=\int \frac{1}{u} d u$

We know that $\int \frac{1}{x} d x=\log |x|+c$

$\Rightarrow \int \frac{1}{u} d u=\log |u|=\log |x+5|$

Then,

$\Rightarrow \int \frac{7 x+10}{x^{2}+7 x+10} d x=\frac{7}{2} \int \frac{2 x+7}{x^{2}+7 x+10} d x-\frac{29}{2} \int \frac{1}{x^{2}+7 x+10} d x$

$=\frac{7}{2}\left(\log \left|x^{2}+7 x+10\right|\right)-\frac{29}{2}\left(\frac{-\log |x+2|}{3}+\frac{\log |x+5|}{3}\right)$

$=\frac{7 \log \left|x^{2}+7 x+10\right|}{2}+\frac{29 \log |x+2|}{6}-\frac{29 \log |x+5|}{6}$

Then,

$\Rightarrow \int \frac{x^{2}}{x^{2}+7 x+10} d x=-\int \frac{7 x+10}{x^{2}+7 x+10} d x+\int 1 d x$

We know that $\int 1 \mathrm{dx}=\mathrm{x}+\mathrm{c}$

$\begin{aligned} \Rightarrow-\int \frac{7 x+10}{x^{2}+7 x+10} d x+\int 1 d x & \\=& \frac{-7 \log \left|x^{2}+7 x+10\right|}{2}-\frac{29 \log |x+2|}{6}+\frac{29 \log |x+5|}{6}+x+c \end{aligned}$

$=\frac{-7 \log |x+2| \log |x+5|}{2}-\frac{29 \log |x+2|}{6}+\frac{29 \log |x+5|}{6}+x+c$

$=-\frac{25 \log |x+2|}{3}+\frac{4 \log |x+5|}{3}+x+c$

$\therefore I=\int \frac{x^{2}}{x^{2}+7 x+10} d x=-\frac{25 \log |x+2|}{3}+\frac{4 \log |x+5|}{3}+x+c$