Initial value problem for Poisson-Darboux equation

Question

This is a question from 'Introduction to Partial Differential Equations' by Peter J. Olver.
The Poisson-Dabroux equation is $$u_{tt}-u_{xx}-\frac{2}{x}u_{x}=0$$
Solve the initial value problem $u(0,x)=0 , u_{t}(0,x)=g(x)$ where $g(x)=g(-x)$. Hint: Set $w=xu$
I used the hint to get the following
$$ w_{xx}= xu_{xx}+2u_{x}$$
$$w_{tt}=xu_{tt}$$
$$w_{t}=xu_{t}$$
Using the given equation to get $$xu_{tt}-xu_{xx}-2u_{x}=0$$
Using the w values obtained before and plugging them into the above equation gives:
$$w_{tt}=w_{xx}$$
Using D'Alembert's formula for a general solution:
$$w(t,x)=\int_{x-t}^{x+t} s g(s) ds$$ Now this is where I'm stuck. I tried using integration by parts but I ended up with a more complicated integral. Also, I reckon the fact that $g$ is even is used somewhere in the calculation of this integral but I can't figure out where either. The solution to this question is not in the solution manual either. Any help appreciated. Thank you!

Answer 1

You're already done. There's nothing wrong with having the solution as an integral. There's no indication that you need to solve it

$$ u(x,t) = \frac{1}{x} \int_{x-t}^{x+t}sg(s) ds $$