Question

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

Answer 1

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

$=\int \sqrt{\frac{(9+x^2)}{9}}dx$

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

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

using $\int \sqrt{(x^2+a^2)}.dx$

$=\frac{1}{2}x\sqrt{(x^2+a^2)}+\frac{a^2}{2}\log |x+\sqrt{(x^2+a^2)}|$

$=\frac{1}{3}[\frac{x}{2}\sqrt{x^2+(3{)}^2}+\frac{(3{)}^2}{2}\log |x+\sqrt{x^2+(3{)}^2}|]$

$=\frac{1}{3}[\frac{x}{2}\sqrt{x^2+9}+\frac{9}{2}\log |x+\sqrt{x^2+9}|+C]$

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

Where $C$ is constant of integration