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CUET · MATHS · PYQ PAPER 2025

For \(x \neq-1\), if \(\int \frac{x e^x d x}{(1+x)^2}=\frac{a e^x}{(1+x)^b}+c\), where \(a, b\) are fixed numbers and \(c\) is the integration constant, then \(a+b\) is equal to

  1. A \(0\)
  2. B 1
  3. C 2
  4. D 3
Verified Solution

Answer & Solution

Correct Answer

(C) 2

Step-by-step Solution

Detailed explanation

\(\int \frac{x e^x d x}{(1+x)^2} = \int \frac{(1+x-1) e^x d x}{(1+x)^2}\) \(= \int \left( \frac{e^x}{1+x} - \frac{e^x}{(1+x)^2} \right) d x\) \(= \int e^x \left( \frac{1}{1+x} + \left(-\frac{1}{(1+x)^2}\right) \right) d x\) \(= \frac{e^x}{1+x} + c\) Comparing with…
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