Question

Find $\int \frac{dx}{x^2+a^2}$ and hence evaluate $\int \frac{dx}{x^2-6x+13}$

Answer 1

Substituting $u=\frac{x}{a}$ in the integral $\int \frac{dx}{x^2+a^2}$. Hence it can be written as $\frac{1}{a}\int \frac{du}{u^2+1}$

=$\frac{1}{a}ta{n}^{-1}u$.

$\int \frac{du}{u^2+1}=ta{n}^{-1}u$ (Standard integral formula)

$\int \frac{dx}{x^2+a^2}=\frac{1}{a}ta{n}^{-1}\frac{x}{a}+C$.

$\int \frac{dx}{x^2-6x+13}=\int \frac{dx}{(x-3{)}^2+4}$

substitute $x-3=u$, then the integral reduces to $\int \frac{du}{u^2+4}$

By applying the formula we obtained for first integral of the question, $\int \frac{du}{u^2+4}=\frac{1}{2}ta{n}^{-1}\frac{u}{2}$

Substitute $u=x-3$ , Integral= $\frac{1}{2}ta{n}^{-1}\frac{x-3}{2}+C_1$