Question

Integrate with respect to $x$:
(i) $\frac{1}{x^2+4}$ (ii) $\frac{1}{x^2+5}$

Answer 1

(i) $\frac{1}{x^2+4}$

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

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

$\therefore \int \frac{dx}{x^2+4}=\frac{1}{2}{\tan }^{-1}\frac{x}{2}+C$.

(ii) $\frac{1}{x^2+5}$

$\int \frac{dx}{x^2+5}=\int \frac{dx}{x^2+(\sqrt{5}{)}^2}$

$=\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\frac{x}{\sqrt{5}}\right)+C$

$\therefore \int \frac{dx}{x^2+5}=\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\frac{x}{\sqrt{5}}\right)+C$.