Tough Integral problem

how to integrate this?

integratal: x/((x-1)^2*(x^2+2x+2))

x /(x - 1)^2* (x^2 + 2x + 2)

first split into partial fractions

x /(x - 1)^2* (x^2 + 2x + 2) = A/(x - 1) + B/(x-1)^2 + (Cx + D)/(x^2 + 2x + 2)

x = A(x - 1)(x^2 + 2x + 2) + B(x^2 +2x + 2) + (Cx+D)(x-1)^2

x = A(x^3 + x^2 - 2) + B(x^2 + 2x + 2) + C(x^3 - 2x^2 + x) + D(x^2 - 2x + 1)

x = x^3(A + C) + x^2(A + B - 2C + D) + x(2B + C - 2D) - 2A + 2B + D

A + C = 0 ==> A = -C
A + B - 2C + D = 0 ==> 3A + B + D = 0 ------(1)==> 6A + 2B + 2D = 0 ------(2)
2B + C - 2D = 1 ==> 2B - A - 2D = 1 --------(3)
-2A + 2B + D = 0---------(4)
Add (1) and (2)
5A + 4B = 1 --------(5)
subtract (1) from(4)
-5A + B = 0 --------(6)
add (5) and (6)
5B = 1 ==> B = 1/5
A = 1/25
C = -1/25
D = -8/25

x /(x - 1)^2* (x^2 + 2x + 2) = (1/25)(1/(x - 1)+(1/5)1/(x-1)^2 -(1/25)x/(x^2 + 2x + 2)-(8/25)1/(x^2+2x+2)

integrate separately

1/25 ∫dx(x - 1) = (1/25) ln I x - 1 I

1/5∫ dx/(x - 1)^2 = - (1/5)(1/(x - 1))

-1/25∫x dx /(x^2 + 2x + 2) = -1/50 ∫ (2x + 2 - 2) dx /(x^2+2x+2)
= -1/50∫(2x + 2) dx /(x^2 + 2x + 2) +1/25∫ dx/(x^2+2x+2)
= -1/50 ln I x^2 + 2x + 2 I + 1/25∫ dx/(x^2 + 2x + 2)

you can add 1/25∫ dx/(x^2 + 2x + 2) to last one -8/25∫ dx/(x^2 + 2x + 2)

so -7/25∫ dx/(x^2 + 2x + 2) = -7/25∫ dx/[(x + 1)^2 + 1] = -(7/25)tan^-1(x + 1)

So the final answer is

(1/25) ln I x - 1 I - (1/5)1/(x - 1) - (1/50) ln I x^2 + 2x + 2 I - (7/25)tan^-1(x + 1) + C

i believe u need to do partial fractions. Go on here:
http://www.twiddla.com/527162
I will attempt it there. Don't worry it is not a spam site, it is just a whiteboard and it makes easier for me to write math problems and u can see my work and chat at the same time.
EDIT:
You are still not online. You don't need to register for it, just copy paste that link, and u are not feeling safe, then google: what is twiddla? I don't have time to wait for u the whole day or type the whole integral out with the stupid text editor that Y!A has to offer
EDIT again:
Oh well, let me give u some steps here:
1) Use partial fractions decomposition that will give you:
Integral of (x/((x-1)^2*(x^2+2x+2)),x) = (Ax+B)/(x^2+2x+2) + C/(x-1) + D/(x-1)^2. Then solve for A, B, C and D.
I got a great idea actually.
There you go:
http://www.wolframalpha.com/input/?i=Integrate+%28x%2F%28%28x-1%29%5E2%2A%28x%5E2%2B2x%2B2%29%29%2Cx%29
Follow that link to wolfram alpha and click on show steps to see the in-between steps. Saves all of us some time!

int(x/((x-1)^2*(x^2+2*x+2)), x) =

(1/25)*ln(x - 1) - 1/(5*(x - 1)) - (1/50)*ln(x^2 + 2*x + 2) - (7/25)*arctan(1 + x) + C