Frobenius method, reduction of order technique

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)c_{n+1}x^{n} + $\displaystyle \sum_{0}^{\infty}$(n+1)c_{n+1}x^{n} - $\displaystyle \sum_{1}^{\infty}$c_{n-1}x^{n}= 0

[n(n+1) + (n+1)]c_{n+1} = c_{n-1}

c_{n+1} = c_{n-1}/ n(n+1) + (n+1)

c_{1} = 0

c_{2} = c_{0}/2^{2 }c_{3} = c_{1}/3^{2 }= 0

c4 = c_{2}/2^{2}*4^{2
c}_{5}= 0

y_{1}(x) = 1 + x^{2}/2^{2} + x^{4}/2^{2}*4^{2} + x^{6}/2^{2}*4^{2}*6^{2} + x^{8}/2^{2}*4^{2}*6^{2}*8^{2} + ...

OK, and now to find y_{2}, I'm kind of stuck. Since in the indicial equation, the roots are both r=0, so the equation I should use is: y_{2} = y_{1}*ln(x) + x^{r1+1}$\displaystyle \sum_{0}^{\infty}$b_{n}x^{n}.

The solution manual of my text book starts off with this:

y_{2} = y_{1} $\displaystyle \int_{}^{}$x^{-1} * (1 + x^{2}/2^{2} + x^{4}/2^{2}*4^{2} + x^{6}/2^{2}*4^{2}*6^{2} + x^{8}/2^{2}*4^{2}*6^{2}*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.