AP EAMCET · Maths · Definite Integration
\(\int_0^1 \frac{x e^x}{(x+1)^2} d x\) is equal to
- A \(\frac{e}{2}\)
- B \(\frac{e}{2}-1\)
- C \(\frac{e}{2}+1\)
- D 2e
Answer & Solution
Correct Answer
(B) \(\frac{e}{2}-1\)
Step-by-step Solution
Detailed explanation
\(I=\int_0^1 \frac{x e^x}{(x+1)^2} d x\) \[ \begin{aligned} I & =\int_0^1 e^x\left[\frac{x}{(x+1)^2}\right] d x \\ & =\int_0^1 e^x\left[\frac{1}{x+1}-\frac{1}{(x+1)^2}\right] d x \end{aligned} \] Let \(f(x)=\frac{1}{x+1} \Rightarrow f^{\prime}(x)=\frac{-1}{(x+1)^2}\)…
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