Question

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

Answer 1

Let $I=\int \sqrt{1+\frac{x^2}{9}}dx=\frac{1}{3}\int \sqrt{9+x^2}dx=\frac{1}{3}\sqrt{(3{)}^2+x^2}dx$
We know that, $\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log |x+\sqrt{x^2+a^2}|+C$
$\therefore I=\frac{1}{3}[\frac{x}{2}\sqrt{x^2+9}+\frac{9}{2}\log |x+\sqrt{x^2+9}|]+C$
$=\frac{x}{6}\sqrt{x^2+9}+\frac{3}{2}\log |x+\sqrt{x^2+9}|+C$