Q60 / 150 · MHT CET 2020 (15 Oct Shift 2) · Mathematics

$\int \mathrm{e}^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} \mathrm{~d} x=$

  • Correct answer
    A
    $\mathrm{e}^{x}\left(\frac{1}{1+x^{2}}\right)+\mathrm{C}$
  • B
    $\mathrm{e}^{x}\left(\frac{-1}{1+x^{2}}\right)+\mathrm{C}$
  • C
    $\mathrm{e}^{x}\left(\frac{2}{1+x^{2}}\right)+\mathrm{C}$
  • D
    $\mathrm{e}^{x}\left(\frac{-2}{1+x^{2}}\right)+\mathrm{C}$

Solution

$\begin{aligned} I &=\int e^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} d x \\ &=\int e^{x} \frac{\left(1+x^{2}-2 x\right)}{\left(1+x^{2}\right)^{2}} d x=\int e^{x}\left[\frac{1+x^{2}}{\left(1+x^{2}\right)^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right] d x \\ &=\int e^{x}\left[\frac{1}{1+x^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right] d x=\frac{e^{x}}{1+x^{2}}+C \end{aligned}$

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

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