CUET · MATHS · PYQ PAPER 2025
\(\int e^x\left(\frac{1-x}{1+x^2}\right)^2\) \(dx\) is equal to
- A \(\frac{e^x}{1+x^2}+C\) : C is an arbitrary constant
- B \(\frac{e^x}{\left(1+x^2\right)^2}+C\) : C is an arbitrary constant
- C \(\frac{-e^x}{\left(1+x^2\right)^2}+C\) :C is an arbitrary constant
- D \(\frac{-e^x}{1+x^2}+C\) : C is an arbitrary constant
Answer & Solution
Correct Answer
(A) \(\frac{e^x}{1+x^2}+C\) : C is an arbitrary constant
Step-by-step Solution
Detailed explanation
\(\int e^x\left(\frac{1-x}{1+x^2}\right)^2 dx = \int e^x\frac{1-2x+x^2}{\left(1+x^2\right)^2} dx\) \(= \int e^x\left(\frac{1+x^2}{\left(1+x^2\right)^2} - \frac{2x}{\left(1+x^2\right)^2}\right) dx\) \(= \int e^x\left(\frac{1}{1+x^2} + \frac{-2x}{\left(1+x^2\right)^2}\right) dx\)…
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