Help in Calculus?

Question

Can anyone help me to solve the following integral by parts. I have no idea how to start.

xe^x / (x+1)^2

Answer 1

Put u = xe^x and dv = 1/(x+1)^2. Then, du = e^x (x +1) and v = -1/(x +1). Then, by parts,


Int u dv = uv - Int v du = -x e^x/(x +1) - Int (- 1/(x+1)) e^x (x +1) = -x e^x/ (x +1) + Int e^x dx = -x e^x/ (x +1) + e^x + C

Not so hard as it looked at first glance

Answer 2

Wow. By parts? Ouch. Try your f(x) to be (x+1)^2, so that you can derive the squared away, so then your f'(x) will be 2(x+1). So that leaves your g'(x) to be xe^x. So, do by parts with that, too. Let's do h(x) and j(x). So your h(x) will be x, so your h'(x) will be 1 and your g'(x) = e^x, so g(x) = e^x. Then combine it all together.

I was taught f(x), f'(x), g(x), and g'(x), but some were taught u, du, v, and dv, which are exactly the same thing, just a different way to write it. So if my f(x) thing isn't what you use, f's are u's and g's are v's. Then f'(x) is du, and g'(x) is dv.

So I just saw the guy's above mine, and I think mine may still work. Just try both of them and see if you get the same answer. I know it's more work, but if you have two differen ways to tackle the same problem, you'll be better off in the long run.

Answer 3

let u = xe^x with dv = (x+1)^(-2)
then du = (1+x)e^x dx and v = - (x+1)^(-1)

so int(u dv) = - xe^x / (x+1) + int( e^x dx)

Nice one