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Calculus Level 2

∫ x e x ( x + 1 ) 2 d x = ? \large \int \dfrac{xe^x}{(x+1)^2} dx = ?

Notation: C C denotes the constant of integration .

x e x x + 1 + C \frac{xe^x}{x+1} + C x e x + C xe^x+C e x x + 1 + C \frac{e^x}{x+1} + C e x ln ⁡ ( x + 1 ) + C e^x \ln(x+1)+C

Nazmus Sakib by Nazmus sakib , 24, Bangladesh

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2 solutions

I = ∫ x e x ( x + 1 ) 2 d x By integration by parts = − x e x x + 1 + ∫ e x + x e x x + 1 d x = − x e x x + 1 + ∫ e x d x = − x e x x + 1 + e x + C = e x x + 1 + C \begin{aligned} I & = \int \frac {xe^x}{(x+1)^2} dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = - \frac {xe^x}{x+1} + \int \frac {e^x + xe^x}{x+1} dx \\ & = - \frac {xe^x}{x+1} + \int e^x dx \\ & = - \frac {xe^x}{x+1} + e^x + C \\ & = \boxed{\dfrac {e^x}{x+1}+C}\end{aligned}

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Akshat Sharda
Sep 3, 2017

∫ e x x + 1 − 1 ( x + 1 ) 2 d x = ∫ e x ( 1 x + 1 + − 1 ( x + 1 ) 2 ) d x \int e^x \frac{ x+ 1-1 }{(x+1)^2}dx=\int e^x \left( \frac{1}{x+1}+\frac{-1}{(x+1)^2} \right)dx

As d d x ( e x f ( x ) ) = e x ( f ( x ) + f ′ ( x ) ) \frac{d}{dx} (e^x f(x)) = e^x (f(x)+f'(x)) , therefore,

∫ e x ( 1 x + 1 + − 1 ( x + 1 ) 2 ) d x = e x x + 1 + C \int e^x \left( \frac{1}{x+1}+\frac{-1}{(x+1)^2} \right)dx = \frac{e^x}{x+1} + C

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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