# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{(x-1)^{2}}{x^{2}+2 x+2} d x$

Solution:

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

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

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

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

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

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

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

$=\int \frac{1}{2 u} d u$

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

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

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

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

Let $u=x+1 \rightarrow d x=d u$

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

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

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

Then,

$\Rightarrow \int \frac{4 x+1}{x^{2}+2 x+2} d x=4 \int \frac{x+1}{x^{2}+2 x+2} d x-3 \int \frac{1}{x^{2}+2 x+2} d x$

$=4\left(\frac{\log \left|x^{2}+2 x+2\right|}{2}\right)-3\left(\tan ^{-1}(x+1)\right)$

$=2 \log \left|x^{2}+2 x+2\right|-3 \tan ^{-1}(x+1)$

Then,

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

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

$\Rightarrow-\int \frac{4 x+1}{x^{2}+2 x+2} d x+\int 1 d x=-2 \log \left|x^{2}+2 x+2\right|+3 \tan ^{-1}(x+1)+x+c$

$\therefore I=\int \frac{(x-1)^{2}}{x^{2}+2 x+2} d x=-2 \log \left|x^{2}+2 x+2\right|+3 \tan ^{-1}(x+1)+x+c$