CUET · MATHS · PYQ PAPER 2023
If \(\int_0^1 \frac{e^x}{1+x} d x=K\), then \(\int_0^1 \frac{e^x}{(1+x)^2} d x\) is equal to:
- A \(K-1+\frac{e}{2}\)
- B \(K+1-\frac{e}{2}\)
- C \(K-1-\frac{e}{2}\)
- D \(K+1+\frac{e}{2}\)
Answer & Solution
Correct Answer
(B) \(K+1-\frac{e}{2}\)
Step-by-step Solution
Detailed explanation
\(\int_0^1 \frac{e^x}{(1+x)^2} d x = \left[ -\frac{e^x}{1+x} \right]_0^1 - \int_0^1 \left(-\frac{1}{1+x}\right) e^x d x\) \(= \left( -\frac{e}{2} - (-\frac{1}{1}) \right) + \int_0^1 \frac{e^x}{1+x} d x\) \(= -\frac{e}{2} + 1 + K\) \(= K + 1 - \frac{e}{2}\)
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