Question

$\int \frac{1}{(x^2+9)\sqrt{x^2-9}}dx$ is equal to (where $C$ is integration constant)

• A. $\frac{1}{9\sqrt{2}}\ln \left|\frac{x\sqrt{2}+\sqrt{x^2-9}}{x\sqrt{2}-\sqrt{x^2-9}}\right|+C$
• B. $\frac{1}{18\sqrt{2}}\ln \left|\frac{x\sqrt{2}-\sqrt{x^2-9}}{x\sqrt{2}+\sqrt{x^2-9}}\right|+C$
• C. $\frac{1}{9\sqrt{2}}\ln \left|\frac{x\sqrt{2}-\sqrt{x^2-9}}{x\sqrt{2}+\sqrt{x^2-9}}\right|+C$
• D. $\frac{1}{18\sqrt{2}}\ln \left|\frac{x\sqrt{2}+\sqrt{x^2-9}}{x\sqrt{2}-\sqrt{x^2-9}}\right|+C$

Answer 1

The correct option is D $\frac{1}{18\sqrt{2}}\ln \left|\frac{x\sqrt{2}+\sqrt{x^2-9}}{x\sqrt{2}-\sqrt{x^2-9}}\right|+C$

$I=\int \frac{1}{(x^2+9)\sqrt{x^2-9}}dx$
Put $x=\frac{1}{t}$
$\Rightarrow dx=-\frac{1}{t^2}dt$
$\Rightarrow I=\int \frac{1}{(\frac{1}{t^2}+9)\sqrt{\frac{1}{t^2}-9}}(-\frac{1}{t^2})dt=\int \frac{-tdt}{(1+9t^2)\sqrt{1-9t^2}}$
Put $1-9t^2=y^2$
$\Rightarrow -9tdt=ydy\Rightarrow I=\int \frac{\frac{y}{9}}{(2-y^2)\sqrt{y^2}}dy=\frac{1}{9}\int \frac{dy}{{\left(\sqrt{2}\right)}^2-y^2}\{\because \int \frac{1}{a^2-y^2}dx=\frac{1}{2a}\ln \left|\frac{a+y}{a-y}\right|+C\}\Rightarrow I=\frac{1}{18\sqrt{2}}\ln \left|\frac{\sqrt{2}+y}{\sqrt{2}-y}\right|+C=\frac{1}{18\sqrt{2}}\ln \left|\frac{\sqrt{2}+\sqrt{1-9t^2}}{\sqrt{2}-\sqrt{1-9t^2}}\right|+C=\frac{1}{18\sqrt{2}}\ln \left|\frac{\sqrt{2}+\sqrt{1-9{\left(\frac{1}{x}\right)}^2}}{\sqrt{2}-\sqrt{1-9{\left(\frac{1}{x}\right)}^2}}\right|+C=\frac{1}{18\sqrt{2}}\ln \left|\frac{x\sqrt{2}+\sqrt{x^2-9}}{x\sqrt{2}-\sqrt{x^2-9}}\right|+C$