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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS1 Mark
MCQ
Choose the correct option from given four options:
$\int\text{e}^\text{x}\Big(\frac{1-\text{x}}{1+\text{x}^2}\Big)^2\text{dx}$ is equal to:
  • ✓
    $\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{C}$
  • B
    $\frac{-\text{e}^\text{x}}{1+\text{x}^2}+\text{C}$
  • C
    $\frac{\text{e}^\text{x}}{(1+\text{e}^2)^2}+\text{C}$
  • D
    $\frac{-\text{e}^\text{x}}{(1+\text{x}^2)^2}+\text{C}$
✓

Answer

Correct option: A.
$\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{C}$
$\int\text{e}^\text{x}\Big(\frac{1-\text{x}}{1+\text{x}^2}\Big)^2\text{dx}$
$=\int\text{e}^\text{x}\frac{1+\text{x}^2-2\text{x}}{(1+\text{x}^2)^2}\text{dx}$
$=\int\text{e}^\text{x}\Big[\frac{1}{(1+\text{x}^2)}-\frac{2\text{x}}{(1+\text{x}^2)^2}\Big]\text{dx}$
$=\int\text{e}^\text{x}[\text{f(x)}+\text{f}'(\text{x})]\text{dx},$ where $\text{f(x)}=\frac{1}{1+\text{x}^2}$
$=\text{e}^\text{x}\text{f(x)}+\text{C}=\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{C}$

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