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,\(\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log\mid x+\sqrt{x^2+a^2}\mid+ C\)

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

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