Integ prob but I'm stuck

  • Thread starter anderma8
  • Start date
  • #1
35
0
I'm trying to do the following:

int (3x^3 + 4x)/(x^2+1)^2 dx

I let u = x^2+1 and I eventually get:

int 3(u-1)+4/u^2 du/2 When I further break this down, I get:

1/2 int 3u^-1 -u^-2 du am I on the right track? When I integrate this, I'm thinking that I have to be doing something wrong...

Any suggestions?
 

Answers and Replies

  • #2
Kurdt
Staff Emeritus
Science Advisor
Gold Member
4,826
6
It looks fine to me as far as the final line there is a slight mistake with the signage.
 
  • #3
35
0
thanks for the 2nd set of eyes... Maybe it should read:

1/2 int (3u^-1) - (u^-2) du

I end up with 1/2 u^-1 +c or 1/2(x^2 + 1) + c since u=x^2+1
 
  • #4
Kurdt
Staff Emeritus
Science Advisor
Gold Member
4,826
6
You have:

[tex]\int \frac{3x^3+4x}{(x^2+1)^2}dx[/tex]

Put [tex] u=x^2+1 \Rightarrow du=2xdx[/tex]

Which gives:

[tex]\frac{3}{2} \int \frac{du}{u} - \frac{3}{2} \int \frac{du}{u^2} + 2 \int \frac{du}{u^2} [/tex]

If you simplify that it will give you something slightly different to what you have. Its basically just a change of sign. Then all you need to do is evaluate the integrals.
 

Related Threads on Integ prob but I'm stuck

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
1K
Replies
14
Views
4K
Replies
5
Views
2K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
5
Views
2K
Replies
10
Views
2K
Replies
10
Views
2K
Top