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GUJCET · Maths · Integrals
\(\int e^x\left(\frac{1-x}{1+x^2}\right)^2 \cdot d x=\underline{\hspace{1cm}} +\mathrm{C}\)
- A \(-\frac{e^x}{1+x^2}\)
- B \(\frac{e^x}{1+x^2}\)
- C \(\frac{e^x}{(1+x^2)^2}\)
- D \(\frac{e^x}{1+x}\)
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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