Question

Find: ${\displaystyle \int \frac{dx}{x^2-6x+13}}$

Answer 1

Given integral is I = ${\displaystyle \int \frac{dx}{x^2-6x+13}}$

= ${\displaystyle \int \frac{dx}{{(x-3)}^2+13-9}}$

= ${\displaystyle \int \frac{dx}{{(x-3)}^2+4}}$

= ${\displaystyle \int \frac{dx}{{(x-3)}^2+2^2}}$

= ${\displaystyle \frac{1}{2}{\tan }^{-1}\left(\frac{x-3}{2}\right)+C}$  ...${\displaystyle [\text{Using}\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1} \frac{x}{a}+C]}$