Question

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

Answer 1

Let $I= \int \sqrt{1+\frac{x^2}{9}}dx=\frac{1}{3}\int \sqrt{9+x^2}dx=\frac{1}{3}\int\sqrt{(3)^2+x^2}\,dx$

It is known that,$\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log\mid x+\sqrt{x^2+a^2}\mid+ C$

∴$I=\frac{1}{3}\bigg[\frac{x}{2}\sqrt{x^2+9}+\frac{9}{2}\log\mid x+\sqrt{x^2+9}\mid\bigg]+C$

=$\frac{x}{6}\sqrt{x^2+9}+\frac{3}{2}\log\mid x+\sqrt{x^2+9}\mid+C$