Question

Solve this step by step.

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Answer 1

★ Solution :

$\sf \: Let Ｉ = ∫ \sf\frac{xeˣ}{(x + 1)²} dx$

$\sf ɪ = ∫ \sf \frac{eˣ}{(x+1)} dx - ∫ \frac{eˣ}{(x+1) {}^{2} } dx$

$\fbox \pink{ \sf \: Applying integration by parts in first integral, we get}$

$\sf \: ɪ = \frac{eˣ}{(x+1)} + ∫ [ \frac{x}{x + 1²} ] \: {e}^{ˣ} dx - ∫ \frac{{e}^{ˣ} }{(x + {1}^{2} )} dx + c$

$\sf \: ɪ = \frac{eˣ}{(x+1)} + ∫ [ \frac{x}{(x+1)²} ] eˣdx - ∫ \frac{eˣ}{(x+1)²} dx + c$

$\boxed{ \sfɪ = \frac{eˣ}{(x+1)} +c}$

Answer 2

Step-by-step explanation:

Given that:

$\tt{ \int \frac{x {e}^{x} }{(x + 1 {)}^{2} }dx }$

What to do:

To evaluate.

Solution:

$\tt{ \int \frac{x {e}^{x} }{(x + 1 {)}^{2} }dx }$

$= \tt{ \int \frac{(x + 1 - 1) {e}^{x} }{(x + 1 {)}^{2} } dx}$

$= \tt{\int {e}^{x} \bigg( \frac{(x - 1)}{x + 1 {)}^{2} } - \frac{1}{(x + 1 {)}^{2} } \bigg)dx}$

$= \tt{\int {e}^{x} \bigg( \frac{1}{(x - 1)} - \frac{1}{(x - 1 {)}^{2} } \bigg)dx}$

$Let \: x + 1 = t⟹dx = dt$

$= \tt{\int {e}^{t - 1} \bigg( \frac{1}{t} - \frac{1}{ {t}^{2} } \bigg)dx}$

$= \tt{ \frac{1}{e} \int {e}^{t} \bigg( \frac{1}{t} - \frac{1}{ {t}^{2} } \bigg)dx}$

Using

$\int {e}^{x}f[(x) + f'(x)]dx = {e}^{x}f(x) + c$

$= \tt{ \frac{ {e}^{t} }{e} \frac{1}{t} + c}$

$= \tt{\frac{ {e}^{t - 1} }{t} + c}$

$= \tt{{e}^{x} \frac{1}{x + 1} + c \: \: \red{Ans.} }$

I hope it's help you...☺