The initial question I have is to evaluate the following indefinite integral:

$\displaystyle \int \frac{x^2-3x+7}{(x^2-4x+6)^2}dx$

(from: calculus, early transcendentals textbook by James Stewart 6th custom edition for MA110. Question from: sec. 7.4 question #37)

The question seems to require some completing the square of the denominator, substituition, and partial fraction integration. I thought I was on the right track although I have been trying for about 5 hours now and I can't get the same answer as the book which is really frustrating.

The answer according to my book is:

$\displaystyle \frac{7}{8}\sqrt{2}\arctan(\frac{x-2}{\sqrt{2}})+\frac{3x-8}{4(x^2-4x+6)}+ constant$

If anyone has some time then I would really appreciate the help.

NOTE: The step I got to and think I am right is:

$\displaystyle Integral = \int\frac{1}{u^2+2}du+\int\frac{u}{(u^2+2)^2}du+\i nt\frac{3}{(u^2+2)^2}du$

This is just after i finished making the integral into partial fractions. Also $\displaystyle u=x-2$ if you were wondering what I used as my substitution.

So ya I got a little stuck here. I have tried finding the integrals seperately although I can't seem to get it to work and it is hard to see where my errors are because the integrals are all seperate and I can't know if even 1 is right. My work so far may also be wrong too though

Thanks in advance for your help,

Luke