So this is the problem:
xy'' + y'- xy = 0

Find the first nonzero terms in a Frobenius series solution of the given differential equation. Then use the reduction of order technique to find the logarithmic term and the first three nonzero terms in a second linearly independent solutions.

I did the first part:
$\displaystyle \sum_{0}^{\infty}$n(n+1)cn+1xn + $\displaystyle \sum_{0}^{\infty}$(n+1)cn+1xn - $\displaystyle \sum_{1}^{\infty}$cn-1xn​= 0
[n(n+1) + (n+1)]cn+1 = cn-1
cn+1 = cn-1/ n(n+1) + (n+1)
c1 = 0
c2 = c0/22

c3 = c1/32 = 0
c4 = c2/22*42
5= 0

y1(x) = 1 + x2/22 + x4/22*42 + x6/22*42*62 + x8/22*42*62*8​2​ + ...

OK, and now to find y2, I'm kind of stuck. Since in the indicial equation, the roots are both r=0, so the equation I should use is: y2 = y1*ln(x) + xr1+1$\displaystyle \sum_{0}^{\infty}$bnxn.

The solution manual of my text book starts off with this:
y2 = y1 $\displaystyle \int_{}^{}$x-1 * (1 + x2/22 + x4/22*42 + x6/22*42*62 + x8/22*42*62*8​2)-2 dx

... I have NO clue where that came from. Someone please explain how to solve this? I'm studying for an exam and an explanation would be immensely helpful.