Question

Find the integral of $\frac{1}{x^2+a^2}$ with respect to $x$ and hence find $\int \frac{1}{x^2-6x+13}dx$.

Answer 1

$\int \frac{1}{x^2+a^2}dx=\int \frac{a{\sec }^2\theta d\theta }{a^2\tan \theta +a^2}$ Put $x=a\tan \theta dx=a{\sec }^2\theta d\theta$
$=\int \frac{a{\sec }^2\theta d\theta }{a^2(1+{\tan }^2\theta )}=\int \frac{{\sec }^2\theta d\theta }{a{\sec }^2\theta }=\frac{1}{a}\int d\theta =\frac{1}{a}\theta +c$
$=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+c$
Also $\int \frac{1}{x^2-6x+13}dx=\int \frac{dx}{x^2-6x+9+4}=\int \frac{dx}{(x-3{)}^2+2^2}=\int \frac{dt}{t^2+2^2}$ ..... where $t=x-3dt=dx$
$=\frac{1}{2}{\tan }^{-1}\left(\frac{x-3}{2}\right)+C$.