Question

$\int \frac{dx}{(x+1)\sqrt{2x-3}}$ equals to (where $C$ is integration constant)

• A. $\frac{2}{\sqrt{5}}{\tan }^{-1}\left(\sqrt{\frac{2x-3}{5}}\right)+C$
• B. $\frac{2}{\sqrt{5}}{\tan }^{-1}\left(\sqrt{2x-3}\right)+C$
• C. $\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\sqrt{\frac{2x-3}{2}}\right)+C$
• D. $\frac{-2}{\sqrt{5}}{\tan }^{-1}\left(\sqrt{\frac{2x-3}{5}}\right)+C$

Answer 1

The correct option is A $\frac{2}{\sqrt{5}}{\tan }^{-1}\left(\sqrt{\frac{2x-3}{5}}\right)+C$

Let $I=\int \frac{dx}{(x+1)\sqrt{2x-3}}$
Putting $2x-3=t^2\Rightarrow dx=tdt$
we get,
$I=\int \frac{tdt}{(\frac{t^2+3}{2}+1)t}=\int \frac{2dt}{t^2+5}$
$=\frac{2}{\sqrt{5}}{\tan }^{-1}\left(\frac{t}{\sqrt{5}}\right)+C\{\because \int \frac{1}{t^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{t}{a}\right)+C\}$
$=\frac{2}{\sqrt{5}}{\tan }^{-1}\sqrt{\frac{2x-3}{5}}+C$