COMEDK · Maths · 28. Indefinite Integration
If \(\int \frac{x e^{x}}{(1+x)^{2}} d x=e^{x} f(x)+c\), then \(f(x)\) is equal to
- A \(\frac{1}{(1+x)^{2}}\)
- B \(\frac{x}{(1+x)}\)
- C \(\frac{1}{1+x}\)
- D \(\frac{x}{(1+x)^{2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{1+x}\)
Step-by-step Solution
Detailed explanation
We have, \(I=\int \frac{x e^{x}}{(1+x)^{2}} d x\) \[ =\int \frac{(x+1-1)}{(x+1)^{2}} e^{x} d x \] \[ \begin{aligned} &=\int\left[\frac{1}{x+1}-\frac{1}{(x+1)^{2}}\right] e^{x} d x \\ &=e^{x} \frac{1}{(1+x)}+C \end{aligned} \] On comparing, we get \[ f(x)=\frac{1}{1+x} \]
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