Q82 / 180 · COMEDK 2023 (Memory Based) · Mathematics

$\int \frac{x d x}{2(1+x)^{3 / 2}}$ is equal to

  • Correct answer
    A
    $\frac{2+x}{\sqrt{1+x}}+C$
  • B
    $\frac{2+x}{x \sqrt{1+x}}+C$
  • C
    $\frac{x}{\sqrt{1+x}}+C$
  • D
    $-\frac{x}{\sqrt{1+x}}+C$

Solution

Let $I=\int \frac{x d x}{2(1+x)^{3 / 2}}$ On putting, $1+x=t$, we get $d x=d t$ $\begin{aligned} \therefore \quad I & =\int \frac{(t-1) d t}{2 t^{3 / 2}}=\frac{1}{2}\left[\int t^{-1 / 2} d t-\int t^{-3 / 2} d t\right] \\ & =\frac{1}{2}\left[\frac{t^{1 / 2}}{1 / 2}-\frac{t^{-1 / 2}}{-1 / 2}\right]+C \\ & =\frac{1}{2} \times 2\left[\sqrt{t}+\frac{1}{\sqrt{t}}\right]+C \\ & =\frac{t+1}{\sqrt{t}}+C=\frac{x+2}{\sqrt{1+x}}+C \end{aligned}$

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

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