# Integrate the function$\sqrt{1+\frac{x^2}{9}}$

Toolbox:
• $\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac {a^2}{2}log |x+\sqrt{x^2+a^2}|+c$
Given $I=\int \sqrt{1+\frac{x^2}{9}}dx$

$1+\frac{x^2}{9}=\frac{9+x^2}{9}$

$=\frac{1}{9}(9+x^2)$

Therefore $I=\sqrt{\frac{1}{9}(x^2+9)}dx$

$=\frac{1}{3}\sqrt{x^2+9}dx$

Clearly this is of the form $\sqrt{x^2+a^2}=$

$=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}log|x+\sqrt{x^2+a^2}|+c$

Here in the place of x we have x and in the place of a we have 3

Therefore $\frac{1}{3}\int \sqrt{x^2+9} dx=\frac{1}{3}[\frac{x}{2}\sqrt{x^2+9}+\frac{9}{2} log |(x+\sqrt{x^2+9}|+c$

$=\frac{x}{6}\sqrt{x^2+9}+\frac{3}{2}log|x+\sqrt{x^2+9}|+c$