What is the Laplace Inverse Transform of $\ln(s)/(s(s+a))$? I've come across a term $\frac{\ln(s)}{s(s+a)}$ in a transformed differential equation.
I am trying to take the inverse transform of it, but I have no idea how to approach it. 
If anyone knows how to solve this, that would be greatly appreciated.

Let $$F(s)=\ln s\\G(s)=\dfrac{1}{s(s+a)}$$therefore we're gonna find the Inverse Laplace Transform (ILT) of $F(s)G(s)$, also if we denote the ILTs of $F(s)$ and $G(s)$ with $f(t)$ and $g(t)$ respectively the $ILT$ of $F(s)G(s)$ is $f(t)*g(t)$ where $*$ denotes the convolution operator. Also we have$$f(t)=\dfrac{u(t)}{t}\\g(t)=\dfrac{1}{a}(1-e^{-at})u(t)$$where $u(t)$ denotes unit step function. Therefore$$f(t)*g(t)=\int_{-\infty}^{\infty}f(\tau)g(t-\tau)d\tau=\dfrac{1}{a}\int_{0}^{t}\dfrac{1-e^{-a\tau}}{t-\tau}d\tau$$which has no closed form so:$$\dfrac{1}{a}\int_{0}^{t}\dfrac{1-e^{-a\tau}}{t-\tau}d\tau=L^{-1}\left(\dfrac{\ln s}{s(s+a)}\right)$$
