Solving a differential equation with wronskians So I am asked to find a solution to this ODE here and I feel like I am missing something very obvious.
I am asked to find the general solution of:
$x^2y"-3xy'+4y=\frac{x^2}{ln(x)}, y>1$
So I first tried to find the homogeneous solution which was just a cauchy Euler equation:
$x^2y"-3xy'+4y=0$
if I solve that, I get $y_{h}(x)=c_1 x^2 + c_2 x^2 ln(x)$
Then I tried to use variation of parameters to solve the particular solution. I obtain:
$W=\begin{bmatrix}
  x^2 & x^2\ln(x) \\
 2x & x+2x\ln(x) \\
\end{bmatrix}$
The wronskian ends up becoming $W=x^3$ so things worked out really nicely.
If I try to find the particular solution though, I can't integrate one of the integrals.
$Y_p(t)=-y_1\int \frac{y_2g(x)}{W}+y_2\int \frac{y_1g(x)}{W} dx$
$Y_p(t)=-x^2\int x dx +x^2\ln(x)\int \frac{x}{ln(x)} dx$
but the second integral can't be done so either I made a mistake or there is another way to solve this.
I can't even use Laplace transforms since I don't have initial conditions so I am a little lost here...
Thanks!

Yes, you made a mistake.
Note that the coefficient of $y''$ in your differential equation is $x^2$, but you're using a formula intended for a d.e. where the coefficient of $y''$ is $1$.
