Q113 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics

$\int \frac{\left(x+\sqrt{1+x^2}\right)^2}{\sqrt{1+x^2}} d x=$

  • A
    $\frac{x}{\sqrt{1+x^2}}+C$
  • B
    $\log \left|x+\sqrt{1+x^2}\right|+C$
  • C
    $x+\sqrt{1+x^2}+C$
  • Correct answer
    D
    $\frac{\left(x+\sqrt{1+x^2}\right)^2}{2}+C$

Solution

Let $I=\int \frac{\left(x+\sqrt{1+x^2}\right)^2}{\sqrt{1+x^2}} d x$ $\begin{aligned} & \text { put } x+\sqrt{1+x^2}=t \\ & \left(1+\frac{x}{\sqrt{1+x^2}}\right) d x=d t \Rightarrow\left(\frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}\right) d x=d t \\ & \because \quad I=\int t d t=\frac{t^2}{2}+C=\frac{\left(x+\sqrt{1+x^2}\right)^2}{2}+C\end{aligned}$

Hence, the correct answer is option: 'D'.

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