As per the title, I evaluated
$$\int\frac{\log(x+a)}{x}\,dx$$
And wanted to make sure my solution is correct, and if not, where I went wrong in my process. Here is my work.
$$\int\frac{\log(x+a)}{x}\,dx$$
$$=\log(a)\int\frac{1}{x}\,dx+\int\frac{\log\left(\frac{x}{a}+1\right)}{x}\,dx$$
For the second integral, let:
$$\frac{x}{a}=u$$
$$=\log(a)\log(x)+\int\frac{\log(1+u)}{u}\,du$$
$$=\log(a)\log(x)-{Li_{2}}\left(-\frac{x}{a}\right)$$
And so
$$\int\frac{\log(x+a)}{x}\,dx=\log(a)\log(x)-{Li_{2}}\left(-\frac{x}{a}\right)$$
Where $Li_{2}(z)$ is the dilogarithm function, and I used:
$$\int\frac{\log(1+z)}{z}\,dz=-Li_{2}(-z)$$
Is this answer correct?
