Mathematics

Find $$\displaystyle\int { \dfrac { { xe }^{ x } }{ { \left( 1+x \right)  }^{ 2 } } dx } $$ is equal to

B
$$\dfrac{-e^{x}}{(1+x)^{2}}+c$$
C
$$e^{x}(x+1)+c$$
D
$$\dfrac{e^{x}}{1+x^{2}}+c$$

ANSWER

$$\cfrac{1}{1+x}e^x+c$$



SOLUTION
Let $$I=\int{\cfrac{xe^x}{(1+x)^2}dx}$$
 $$I=\int{\cfrac{xe^x+e^x-e^x}{(1+x)^2}dx}$$
 $$I=\int{\cfrac{(x+1)e^x}{(1+x)^2}dx}$$ $$+\int{\cfrac{e^x}{(1+x)^2}dx}$$
$$I=\int{\cfrac{e^x}{(1+x)}dx}$$ $$+\int{\cfrac{e^x}{(1+x)^2}dx}$$
Let $$f(x)=\cfrac{1}{1+x}$$
So, $$g(x)=f'(x)=\cfrac{-1}{(1+x)^2}$$
Using identity
$$I=\int{f(x)e^xdx}+$$$$\int{f'(x)e^xdx}=f(x)e^x$$
So, $$I=\cfrac{1}{1+x}e^x+c$$

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Single Correct Medium Published on 17th 09, 2020
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