Question

Evaluate: ${\int }_0^3\frac{dx}{9+x^2}$

Answer 1

We know that $\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}(\frac{x}{a})$

In given question , the value of $a$ is $3.$

So we get ${\int }_0^3\frac{1}{x^2+3^2}dx$

$=\frac{1}{3}{\tan }^{-1}(\frac{x}{3})$

$=\frac{1}{3}({\tan }^{-1}1-{\tan }^{-1}0)$

$=\frac{1}{3}\left(\frac{\pi }{4}\right)=\frac{\pi }{12}$