Question

Evaluate:
$\int \frac{1}{4+9x^2}dx$

Answer 1

$\int \frac{dx}{4+9x^2}=\frac{1}{9}\int \frac{dx}{(\frac{2}{3}{)}^2+x^2}$

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

$=\frac{1}{a}\times \frac{1}{\left(\frac{2}{3}\right)}ta{n}^{-}\left(\frac{3x}{2}\right)+c$

$=\frac{1}{6}ta{n}^{-1}\left(\frac{3x}{2}\right)+c$