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GUJCET · Maths · Integrals

\(\int e^x\left(\frac{1-x}{1+x^2}\right)^2 \cdot d x=\underline{\hspace{1cm}} +\mathrm{C}\)

  1. A \(-\frac{e^x}{1+x^2}\)
  2. B \(\frac{e^x}{1+x^2}\)
  3. C \(\frac{e^x}{(1+x^2)^2}\)
  4. D \(\frac{e^x}{1+x}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{e^x}{1+x^2}\)

Step-by-step Solution

Detailed explanation

\(\int e^x\left(\frac{1-x}{1+x^2}\right)^2 \, dx = \int e^x\left(\frac{(1-x)^2}{(1+x^2)^2}\right) \, dx\) \(= \int e^x\left(\frac{1-2x+x^2}{(1+x^2)^2}\right) \, dx\)
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