CUET · MATHS · PYQ PAPER 2023
\(\int e^x\left(\frac{1-x}{1+x^2}\right)^2\) dx is equal to :
- A \(\frac{e^x}{1+x^2} + C\)
- B \(-\frac{e^x}{1+x^2}+C\)
- C \(\frac{e^x}{(1+x^2)^2} + C\)
- D \(-\frac{e^x}{\left(1+x^2\right)^2}+C\)
Answer & Solution
Correct Answer
(A) \(\frac{e^x}{1+x^2} + C\)
Step-by-step Solution
Detailed explanation
\(\int e^x\left(\frac{1-x}{1+x^2}\right)^2 dx = \int e^x\left(\frac{1-2x+x^2}{(1+x^2)^2}\right) dx\) \(\qquad = \int e^x\left(\frac{(1+x^2)-2x}{(1+x^2)^2}\right) dx\) \(\qquad = \int e^x\left(\frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2}\right) dx\) \(\qquad = \frac{e^x}{1+x^2} + C\)
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