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Question

Mathematics Question on integral

If ∫xex(1+x)2dx=exf(x)+c,\int \frac{xe^{x}}{\left(1+x\right)^{2}} dx = e^{x} f\left(x\right) +c, then f(x)f(x) is equal to

A

1(1+x)2 \frac{1}{(1 + x)^2}

B

x(1+x) \frac{x}{(1 + x)}

C

1(1+x) \frac{1}{(1 + x)}

D

x(1+x)2 \frac{x}{(1 + x)^2}

Answer

1(1+x) \frac{1}{(1 + x)}

Explanation

Solution

∫xex(1+x)2dx=∫ex(x+1−1)(1+x)2dx\int \frac{xe^{x}}{\left(1+x\right)^{2}} dx =\int e^{x} \frac{\left(x+1-1\right)}{\left(1+x\right)^{2}} dx
=∫ex[1x+1−1(x+1)2]dx= \int e^{x} \left[\frac{1}{x+1} - \frac{1}{\left(x +1\right)^{2}}\right] dx
This is a type of ∫ex(f(x)+f′(x))dx=exf(x)+c\int e^{x} \left(f\left(x\right) +f'\left(x\right)\right)dx =e^{x}f\left(x\right)+c
∴  ∫ex[1x+1−1(x+1)2]dx=ex(1x+1)+c\therefore \, \, \int e^{x} \left[\frac{1}{x+1} - \frac{1}{\left(x+1\right)^{2}}\right] dx =e^{x} \left(\frac{1}{x+1}\right)+c
Hence, f(x)=1x+1f\left(x\right) = \frac{1}{x+1}