1. ## Definite Intergral

Any hints on where to start?

Integral from 1 to 0 of (x*e^x)/(x+1)^2 dx

2. Originally Posted by millerst
Any hints on where to start?

Integral from 1 to 0 of (x*e^x)/(x+1)^2 dx
Make the following observation

$\frac{x\cdot e^x}{(1+x^2)}=\frac{e^x}{1+x}-\frac{e^x}{(1+x)^2}$

3. Or we should say that clearly ,

$\frac{xe^{x}}{(1+x)^2} = \frac{ e^x (x+1) - e^x }{ (1+x)^2}$

Let $D$ be $\frac{d}{dx}$ ( the differential operator )

$= \frac{ e^x (x+1) - e^x }{ (1+x)^2} = \frac{ (x+1) D(e^x) - D(x+1) e^x }{ (1+x)^2 }$

can you see something special ?

4. Originally Posted by simplependulum
Or we should say that clearly ,

$\frac{xe^{x}}{(1+x)^2} = \frac{ e^x (x+1) - e^x }{ (1+x)^2}$

Let $D$ be $\frac{d}{dx}$ ( the differential operator )

$= \frac{ e^x (x+1) - e^x }{ (1+x)^2} = \frac{ (x+1) D(e^x) - D(x+1) e^x }{ (1+x)^2 }$

can you see something special ?
That works too. But I was actually making the observation $\frac{xe^x}{(1+x)^2}=\frac{e^x}{1+x}-\frac{e^x}{(1+x^2}=D\left(e^x\right)\cdot\frac{1}{ 1+x}+e^x\cdot D\left(\frac{1}{1+x}\right)$

Same thing, two different ways.