Question

Evalaute $\int \frac{dx}{(x^2+4)\sqrt{4x^2+1}}$ (where $C$ is integration constant)

• A. $\frac{-1}{4\sqrt{15}}\ln \left|\frac{2\sqrt{4x^2+1}-\sqrt{15}x}{2\sqrt{4x^2+1}+\sqrt{15}x}\right|+C$
• B. $\frac{-1}{4\sqrt{15}}\ln |\sqrt{15}x+\sqrt{4x^2+1}|+C$
• C. $\frac{-1}{4\sqrt{15}}\ln \left|\frac{\sqrt{4x^2+1}-\sqrt{15}x}{\sqrt{4x^2+1}+\sqrt{15}x}\right|+C$
• D. $\frac{-1}{2\sqrt{15}}\ln \left|\frac{\sqrt{4x^2+1}-\sqrt{15}x}{\sqrt{4x^2+1}+\sqrt{15}x}\right|+C$

Answer 1

The correct option is A $\frac{-1}{4\sqrt{15}}\ln \left|\frac{2\sqrt{4x^2+1}-\sqrt{15}x}{2\sqrt{4x^2+1}+\sqrt{15}x}\right|+C$

$I=\int \frac{dx}{(x^2+4)\sqrt{4x^2+1}}$
Put $x=\frac{1}{t}\Rightarrow dx=-\frac{1}{t^2}dt$ , we get
$I=\int \frac{-\frac{1}{t^2}dt}{(\frac{1}{t^2}+4)\sqrt{\frac{4}{t^2}+1}}=\int \frac{-tdt}{(1+4t^2)\sqrt{4+t^2}}$
Again put, $4+t^2=z^2\Rightarrow tdt=zdz$
$\Rightarrow I=\int \frac{-zdz}{(1+4(z^2-4\left)\right)z}$
$=\int \frac{-dz}{4z^2-15}=-\frac{1}{4}\int \frac{dz}{z^2-\frac{15}{4}}=-\frac{1}{4}.\frac{1}{\sqrt{15}}\ln \left|\frac{2z-\sqrt{15}}{2z+\sqrt{15}}\right|+C=-\frac{1}{4\sqrt{15}}\ln \left|\frac{2\left(\sqrt{4+t^2}\right)-\sqrt{15}}{2\left(\sqrt{4+t^2}\right)+\sqrt{15}}\right|=\frac{-1}{4\sqrt{15}}\ln \left|\frac{2\sqrt{4x^2+1}-\sqrt{15}x}{2\sqrt{4x^2+1}+\sqrt{15}x}\right|+C$