Question

Find the integral of $\frac{1}{x^2+a^2}$ w.r.t. x and hence find $\int \frac{1}{3+2x+x^2}dx$

Answer 1

$\int \frac{1}{x^2+a^2}dx$

$=\frac{1}{a^2}\int \frac{1}{\frac{x^2}{a^2}+1}dx$

$=\frac{1}{a^2}\frac{{\tan }^{-1}\left(\frac{x}{a}\right)}{\frac{1}{a}}$

$=\frac{{\tan }^{-1}\left(\frac{x}{a}\right)}{a}$

$\therefore \int \frac{1}{x^2+a^2}dx=\frac{{\tan }^{-1}\left(\frac{x}{a}\right)}{a}$ ....... $\left(1\right)$

Consider, $I=\int \frac{1}{3+2x+x^2}dx$

$=\int \frac{1}{3-1+1+2x+x^2}dx$

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

Put $x+1=u⟹dx=du$

$⟹I=\int \frac{1}{u^2+(\sqrt{2}{)}^2}du$

$=\frac{{\tan }^{-1}\left(\frac{u}{\sqrt{2}}\right)}{\sqrt{2}}$

$=\frac{{\tan }^{-1}\left(\frac{x+1}{\sqrt{2}}\right)}{\sqrt{2}}$