Question

Evaluate $\int \frac{dx}{\left[\left(x+1\right)\left(\sqrt{4x+3}\right)\right]}$

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

Answer 1

The correct option is C

$2{\tan }^{-1}\sqrt{4x+3}+c$

Explanation for the correct option.

Finding the integral value

Given, $\int \frac{dx}{\left[\left(x+1\right)\left(\sqrt{4x+3}\right)\right]}$

Solving the integral,

Let,

$\begin{array}{c}4x+3=z^2\\ \Rightarrow 4dx=2zdz\\ \Rightarrow 2dx=zdz\end{array}$

Now, substituting this in the integral we get,

$\begin{array}{c}\int \frac{dx}{(x+1)\sqrt{4x+3}}=\frac{1}{2}\int \frac{zdz}{z\left(\frac{z^2-3}{4}+1\right)}\\ =\frac{4}{2}\int \frac{dz}{z^2+1}\\ =2\int \frac{dz}{z^2+1}\\ =2{\tan }^{-1}z+c\\ =2{\tan }^{-1}\sqrt{4x+3}+c\left[\because z=\sqrt{4x+3}\right]\end{array}$

Hence, the correct answer is Option (C) .