Question

$\int \frac{dx}{(x+1)\sqrt{4x+3}}$ is equal to (where $C$ is integration constant)

• A. $2{\tan }^{-1}\left(\sqrt{4x+3}\right)+C$
• B. ${\tan }^{-1}\left(\sqrt{4x+3}\right)+C$
• C. ${\tan }^{-1}\left(\sqrt{x+1}\right)+C$
• D. $2{\tan }^{-1}\left(\sqrt{x+1}\right)+C$

Answer 1

The correct option is A $2{\tan }^{-1}\left(\sqrt{4x+3}\right)+C$

$I=\int \frac{dx}{(x+1)\sqrt{4x+3}}$
Let $4x+3=t^2$
$\Rightarrow 2dx=tdt$
$\Rightarrow I=\int \frac{t}{2(\frac{t^2-3}{4}+1)\cdot t}dt=\int \frac{2}{t^2+1}dt=2{\tan }^{-1}t+C\{\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C\}=2{\tan }^{-1}\left(\sqrt{4x+3}\right)+C$