What is the integral of X over e^X.  Please show your work.
X/e^X

x/e^x=xe^-x you can use INT(udv)=uv- INT(vdu)=> u=x dv=e^-x =>
xe^-x=-xe^-x+INT(e^-x)=-xe^-x -e^-x

The other elementery way is to use power series representation.

Ummm...I didn't get that answer at all.

What I got was this...and please tell me if I'm in error...


    e^-x[1+xln(e)] 
_ ____________________
    
        ln(e)^2

This should be a simple integral
x/e^x = x*e^(-x) INT = integral
integrating.. using INT(u*dv) = (u*v)-INT(v*du)
u = x and dv = e^-x
subsituting and simplifying we get

 = -xe^-x + INT(e^-x)dx
 = -xe^-x - e^-x
 = -(e^-x)*(1+x)
 = -(1+x)/e^x

I think this must be the answer ??????
In case of doubt differentiate it to get the Q' back.

Let me join the fun. This reminds me of some my friends during some
discussions. We argue on something---when we are almost all speaking
of the same thing.

The 3 answers shown are all the same answer.
The simplified form is -(x+1) / e^x

-xe^-x -e^-x
= -x(e^(-x)) -e^(-x)
= [-e^(-x)]*[x+1]
= -(x+1) / e^x  
Same.

---------------------
  e^-x[1+xln(e)] 
_ ____________________
    
        ln(e)^2
Since ln(e) = 1, then, your answer is
-{e^-x)[1 +x(1)] / (1)^2}
= -{[e^(-x)]*[1 +x] / 1}
= -{[e^(-x)]*[1 +x]}
= -{(1+x) / e^x}
= -(x+1) / e^x
Same. So you really got the answer also.

-----------
Why -(x+1) / e^x ?
Let me not skip some steps.

Integral of x over e^x.
Actually, that should have been
INT.[x / e^x]dx.
The experts tend to consider the "dx" as always there. But if you are
just a beginner, it pays to always consider the "dx" be there.
Integral of expressions, like (x /e^x), means nothing. The integrand
must always have a differential, a "dx" in this case.

So, INT.[x /e^x]dx
= INT.[x(e^(-x))]dx
= INT.[x *(e^(-x))dx] ----(i)

Let u = x
Then, du = dx

And so dv = (e^(-x))dx
Then, v = INT.[e^(-x)]dx 
= INT.[e^(-x)](-dx)(-1)]
= -INT.[e^(-x)](-dx)
= -[e^(-x)]

Hence,
INT.[x /e^x]dx
= INT.[x(e^(-x))]dx
= INT.[x *(e^(-x))dx] ----(i)
= u*v -INT.[v*du]
= x*(-[e^(-x)]) -INT.[-[e^(-x)]*dx]
= x*(-[e^(-x)]) -INT.{[e^(-x)](-dx)}
= x*(-[e^(-x)]) -{e^(-x)}
= x*(-[e^(-x)]) +[-{e^(-x)}]
= (-[e^(-x)])*(x+1)
= -(x+1) / e^x

---------------
Some experts cringe when somebody tries to "re-invent the wheel". But
it is always a pleasure for me to re-invent the wheel infront of those
who may not know yet how the wheel was invented. Well, the wheel was
invented in many ways. Some of those ways, I know.

My Calculator (TI-89) says :

[-x/ln(e) - 1/(ln(e))^2] * e^-x

[-x/ln(e) - 1/(ln(e))^2] * e^-x
= [-x/1 - 1/(1)^2] * [1 /e^x]
= [-x -1]*[1 /e^x]
= [-(x+1)]*[1 /e^x]
= -(x+1) / e^x

Same.