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Question: Answered & Verified by Expert
$\int \frac{\left(x+\sqrt{1+x^2}\right)^2}{\sqrt{1+x^2}} d x=$
MathematicsIndefinite IntegrationTS EAMCETTS EAMCET 2020 (14 Sep Shift 1)
Options:
  • A $\frac{x}{\sqrt{1+x^2}}+C$
  • B $\log \left|x+\sqrt{1+x^2}\right|+C$
  • C $x+\sqrt{1+x^2}+C$
  • D $\frac{\left(x+\sqrt{1+x^2}\right)^2}{2}+C$
Solution:
2058 Upvotes Verified Answer
The correct answer is: $\frac{\left(x+\sqrt{1+x^2}\right)^2}{2}+C$
Let $I=\int \frac{\left(x+\sqrt{1+x^2}\right)^2}{\sqrt{1+x^2}} d x$
$\begin{aligned} & \text { put } x+\sqrt{1+x^2}=t \\ & \left(1+\frac{x}{\sqrt{1+x^2}}\right) d x=d t \Rightarrow\left(\frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}\right) d x=d t \\ & \because \quad I=\int t d t=\frac{t^2}{2}+C=\frac{\left(x+\sqrt{1+x^2}\right)^2}{2}+C\end{aligned}$

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