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$\int \frac{d x}{(x+1) \sqrt{4 x+3}}$ is equal to
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Verified Answer
The correct answer is:
$2 \tan ^{-1} \sqrt{4 x+3}+c$
Let $\quad I=\int \frac{d x}{(x+1) \sqrt{4 x+3}}$
Put $4 x+3=t^2 \Rightarrow 4 d x=2 t d t$
$\begin{aligned}
\therefore \quad I= & \frac{1}{2} \int \frac{t d t}{\left(\frac{t^2-3}{4}+1\right) t}=2 \int \frac{d t}{1+t^2} \\
& =2 \tan ^{-1} t+c=2 \tan ^{-1} \sqrt{4 x+3}+c
\end{aligned}$
Put $4 x+3=t^2 \Rightarrow 4 d x=2 t d t$
$\begin{aligned}
\therefore \quad I= & \frac{1}{2} \int \frac{t d t}{\left(\frac{t^2-3}{4}+1\right) t}=2 \int \frac{d t}{1+t^2} \\
& =2 \tan ^{-1} t+c=2 \tan ^{-1} \sqrt{4 x+3}+c
\end{aligned}$
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