Question

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

Answer 1

Let $\int \sqrt{1+\frac{x^2}{9}}dx=\int \sqrt{\frac{9+x^2}{9}}dx=\frac{1}{3}\int \sqrt{(3{)}^2+x^2}dx$
$=\frac{1}{3}.\frac{x}{2}\sqrt{(3{)}^2+x^2}+\frac{9}{2\times 3}log|x+\sqrt{9+x^2}|+C[\because \int \sqrt{a^2+x^2}dx=\frac{1}{2}x\sqrt{a^2+x^2}+\frac{1}{2}{a}^{2}log|x+\sqrt{a^2+x^2}|+C]=\frac{x}{6}\sqrt{x^2+9}+\frac{3}{2}log|x+\sqrt{9+x^2}|+C$