# Integ prob but I'm stuck

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?

Kurdt
Staff Emeritus
Gold Member
It looks fine to me as far as the final line there is a slight mistake with the signage.

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

Kurdt
Staff Emeritus
Gold Member
You have:

$$\int \frac{3x^3+4x}{(x^2+1)^2}dx$$

Put $$u=x^2+1 \Rightarrow du=2xdx$$

Which gives:

$$\frac{3}{2} \int \frac{du}{u} - \frac{3}{2} \int \frac{du}{u^2} + 2 \int \frac{du}{u^2}$$

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.