Q60 / 100 · BITSAT 2024 (Memory Based Paper 1) · Mathematics

$\int \sqrt{x+\sqrt{x^2+2}} d x=$

  • A
    $\frac{3}{2}(x+\sqrt{x+2})^{3 / 2}-2\left(x+\sqrt{x^2+2}\right)^{1 / 4}+C$
  • B
    $\frac{1}{3}\left(x+\sqrt{x^2+2}\right)^{3 / 2}-2\left(x+\sqrt{x^2+2}\right)^{1 / 4}+C$
  • C
    $\left(x+\sqrt{x^2+2}\right)^{-3 / 2}-2\left(x+\sqrt{x^2+2}\right)^{-1 / 2}+C$
  • Correct answer
    D
    $\frac{\left(x+\sqrt{x^2+2}\right)^2-6}{3 \sqrt{x+\sqrt{x^2+2}}}+C$

Solution

$\int \sqrt{x+\sqrt{x^2+2}} d x$ $\sqrt{x+\sqrt{x^2+2}}=t \Rightarrow x+\sqrt{x^2+2}=t^2$ $\sqrt{x^2+2}=t^2-x$ $\Rightarrow x^2+2=t^4+x^2-2 t^2 x$ $\Rightarrow \mathrm{x}=\frac{\mathrm{t}^4-2}{2 \mathrm{t}^2} \Rightarrow \mathrm{dx}=\frac{\mathrm{t}^4+2}{\mathrm{t}^3} \mathrm{dt}$ $\int \mathrm{t} \cdot \frac{\mathrm{t}^4+2}{\mathrm{t}^3} \mathrm{dt}=\int\left(\mathrm{t}^2+\frac{2}{\mathrm{t}^2}\right) \mathrm{dt}=\frac{\mathrm{t}^3}{3}-\frac{2}{\mathrm{t}}+\mathrm{C}$ $=\frac{t^3}{3}-\frac{2}{t}+C=\frac{t^4-6}{3 t}+C$ $=\frac{\left(x+\sqrt{x^2+2}\right)^2-6}{3 \sqrt{x+\sqrt{x^2+2}}}+C$

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

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