Question

$\int \frac{1}{4+9x^2}dx$ is equal to
(where $C$ is constant of integration)

• A. $\frac{1}{6}{\tan }^{-1}\frac{3x}{2}+C$
• B. $\frac{1}{3}{\tan }^{-1}\frac{6x}{2}+C$
• C. $\frac{1}{2}{\tan }^{-1}\frac{5x}{2}+C$
• D. $\frac{1}{3}{\tan }^{-1}\frac{7x}{2}+C$

Answer 1

The correct option is A $\frac{1}{6}{\tan }^{-1}\frac{3x}{2}+C$

we have,
$\int \frac{1}{4+9x^2}dx$

$=\frac{1}{9}\int \frac{1}{\frac{4}{9}+x^2}$
$=\frac{1}{9}\int \frac{1}{{\left(\frac{2}{3}\right)}^2+x^2}dx$
$=\frac{1}{9}\cdot \frac{1}{\left(\frac{2}{3}\right)}{\tan }^{-1}\left(\frac{3x}{2}\right)+C$
$=\frac{1}{6}{\tan }^{-1}\left(\frac{3x}{2}\right)+C$