Where did I go wrong with $\int{\frac{\ln{x}}{x + a}dx}$

Question

I tried evaluating this integral over the complexes like the following but when comparing to the Wolfram Alpha answer (including graphically), it gets a different answer. I am not sure where I went wrong, but my theory is that is has to do with the multivalued nature of the complex logarithm. I would appreciate some insight. Assume x is complex and a is non-zero real. $$\begin{align*} \int{\frac{\ln{x}}{x + a}} dx&= \int{\frac{\ln{(u - a)}}{u}}du \qquad \small{u = x + a, x = u - a, du = dx}\\ &= \int{\frac{\ln{((-a)(1 -\frac{u}{a}))}}{u}du}\qquad\text{factor our }-a\\ &= \int{\frac{\ln{(1 -\frac{u}{a} ) + \ln(-a)}}{u}}du\qquad \text{properties of }\ln\\ &= \int{\frac{\ln{(1 -\frac{u}{a} )}}{u}}du + \ln(-a)\int{\frac{1}{u}}du\\ &= \int{\frac{\ln{(1 -\frac{u}{a} )}}{u}}du + \ln(-a)\ln(u) + C \qquad\text{set }v=\frac{u}{a}, u=av, dv=\frac{1}{a}du\\ &= \int{\frac{\ln{(1 -v)}}{v}}dv + \ln(-a)\ln(u) + C \\ &= \textrm{-Li}_2(v) + \ln(-a)\ln(u) + C\qquad\text{with }\mathrm{Li}_2(z)=-\int_0^z\frac{\ln(1-t)}{t}\,dt\\ &= \textrm{-Li}_2\left(\frac{u}{a}\right) + \ln(-a)\ln(u) + C\qquad\left(v=\frac{u}{a}\right)\\ &= \textrm{-Li}_2\left(\frac{x + a}{a}\right) + \ln(-a)\ln(x + a) + C\qquad (u=x+a) \end{align*}$$

Answer 1

Your answer is the function $f$ below, and w-alpha gives the function $g$ below: $$ \begin{aligned} f(x) &= -\operatorname{Li}_2\left(1+ \frac xa\right) + \ln(-a)\ln(x + a) + \text{some constant}\ ,\\ g(x) &= +\operatorname{Li}_2\left(-\frac xa\right) + \ln x \ln\frac {x+a}a + \text{some other constant}\ . \end{aligned} $$ Let now $z$ be $z=-x/a$. Then there is a functional equation satisfied by the dilogarithm, connecting $z$ and $(1-z)$, this is $(5)$ on the mathworld-wolfram page: $$ \operatorname{Li}_2 (z) + \operatorname{Li}_2 (1-z) = -\ln z\ln(1-z) +\text{(known constant)} \ . $$ Using it, $$ \begin{aligned} g(x)-f(x) &= \operatorname{Li}_2(z) + \operatorname{Li}_2(1-z) + \ln(x+a)(\ln x-\ln (-a)) -\ln x\ln a +\text{difference constant} \\ &= -\ln\left(-\frac xa\right) \ln\frac{x+a}a + \ln(x+a)\ln\left(-\frac xa\right) - \ln x\ln a +\text{constant} \\ &= \ln\left(-\frac xa\right) \ln a - \ln x\ln a +\text{constant} \\ &= \ln\left(-\frac 1a\right) \ln a +\text{constant} \\ &=\text{new constant .} \end{aligned} $$