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Question: Answered & Verified by Expert
$\int \frac{d x}{\left(2 a x+x^2\right)^{\frac{3}{2}}}=$
MathematicsIndefinite IntegrationJEE Main
Options:
  • A $\frac{-1}{a^2} \frac{(x+a)}{\sqrt{2 a x+x^2}}+c$
  • B $\frac{-(x+a)}{\sqrt{2 a x+x^2}}+c$
  • C $\frac{1}{2 a^2} \frac{(x+a)}{\sqrt{2 a x+x^2}}+c$
  • D $\frac{-1}{a} \frac{(x+a)}{\sqrt{2 a x+x^2}}+c$
Solution:
2373 Upvotes Verified Answer
The correct answer is: $\frac{-1}{a^2} \frac{(x+a)}{\sqrt{2 a x+x^2}}+c$
We have,
$$
\begin{aligned}
& \int \frac{1}{\left(2 a x+x^2\right)^{3 / 2}} d x=\int \frac{1}{x^3\left(\frac{2 a}{x}+1\right)^{3 / 2}} d x \\
& \text { Put } \begin{aligned}
\frac{2 a}{x}+1 & =t \Rightarrow x=\frac{2 a}{t-1} \Rightarrow \frac{-2 a}{x^2} d x=d t \\
& =-\frac{1}{2 a} \int \frac{t-1}{2 a(t)^{3 / 2}} d t=-\frac{1}{4 a^2} \int \frac{t-1}{t^{3 / 2}} d t
\end{aligned}
\end{aligned}
$$


$$
\begin{aligned}
& =-\frac{1}{4 a^2} \int\left(t^{-1 / 2}-t^{-3 / 2}\right) d t=-\frac{1}{4 a^2}\left[2 t^{1 / 2}+2 t^{-1 / 2}\right]+c \\
& =-\frac{1}{2 a^2}\left(t^{1 / 2}+t^{-1 / 2}\right)+c=-\frac{1}{2 a^2}\left(\frac{t+1}{\sqrt{t}}\right)+c
\end{aligned}
$$

Put $t=\frac{2 a}{x}+1$
$$
\begin{aligned}
& =-\frac{1}{2 a^2}\left(\frac{\frac{2 a}{x}+1+1}{\sqrt{\frac{2 a}{x}+1}}\right)+c=-\frac{1}{a^2}\left(\frac{\frac{a}{x}+1}{\sqrt{\frac{2 a}{x}+1}}\right)+c \\
& =-\frac{1}{a^2}\left(\frac{\frac{a+x}{x}}{\sqrt{\frac{2 a+x}{x}}}\right)+c=-\frac{1}{a^2}\left(\frac{a+x}{\sqrt{2 a x+x^2}}\right)+c
\end{aligned}
$$

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