Evaluate-intdgral-xe-x-x-1-2-dx-please-help-

Question Number 5893 by sanusihammed last updated on 04/Jun/16

E v a l u a t e i n t d g r a l \frac{{x e}^{x}}{{(x + 1)}^{2}} d x

p l e a s e h e l p .

Commented by FilupSmith last updated on 04/Jun/16

integrate by parts

u = {x e}^{x}, d v = \frac{1}{{(x + 1)}^{2}} d x

∴ d u = (x + 1) e^{x} d x, ∴ v = - \frac{1}{x + 1} d x

\int u d v = u v - \int v d u

∴ \int {x e}^{x} \frac{1}{{(x + 1)}^{2}} d x = - \frac{{x e}^{x}}{x + 1} + \int \frac{1}{x + 1} (x + 1) e^{x} d x

= - \frac{{x e}^{x}}{x + 1} + e^{x} + c

= e^{x} - \frac{{x e}^{x}}{x + 1} + c

= e^{x} (1 - \frac{x}{1 + x}) + c

= e^{x} (\frac{(1 + x) - x}{1 + x}) + c

= e^{x} (\frac{1}{1 + x}) + c

= \frac{e^{x}}{1 + x} + c, c = constant

Commented by sanusihammed last updated on 04/Jun/16

T h a n k s a l o t

Answered by Yozzii last updated on 04/Jun/16

\int \frac{{x e}^{x}}{{(x + 1)}^{2}} d x = \int \frac{{x e}^{x} + e^{x} - e^{x}}{{(x + 1)}^{2}} d x

= \int \frac{(x + 1) e^{x} - 1 \times e^{x}}{{(x + 1)}^{2}} d x

= \int \frac{(x + 1) \frac{{d e}^{x}}{d x} - e^{x} \frac{d (x + 1)}{d x}}{{(x + 1)}^{2}} d x

= \int \frac{d}{d x} (\frac{e^{x}}{1 + x}) d x

= \frac{e^{x}}{1 + x} + C

Commented by FilupSmith last updated on 04/Jun/16

oh wow, although both methods are correct,

I really like the simplicity of this method!

Commented by sanusihammed last updated on 04/Jun/16

T h a n k s s o m u c h