Integrating $\frac{\ln{ax}}{x\ln{bx}}$ The question is just to find $$\int\frac{\ln{ax}}{x\ln{bx}}\,dx$$ With $a,b,x>0$.
Now I attempted it using the substitution $u=\ln{x}$, $du=\frac{1}{x}dx$:
$$\int\frac{\ln\left(ax\right)}{x\ln\left(bx\right)}dx=\int\frac{\ln a+\ln x}{x\left(\ln b+\ln x\right)}dx=\int\frac{\ln a+u}{\ln b+u}du
 =\ln a\int\frac{du}{\ln b+u}+\int\frac{u}{\ln b+u}du$$
Now $\int\frac{du}{\ln b+u}$ is just $\ln\left(\ln b+u\right)+C$ and
$$\int\frac{u}{\ln b+u}du=\int\frac{\ln b+u-\ln b}{\ln b+u}du=\int du-\ln b\int\frac{du}{\ln b+u}=u-\ln b\cdot\ln\left(\ln b+u\right)+C$$
 Adding these all up gives us $$\ln a\ln\left(\ln b+u\right)-\ln b\ln\left(\ln b+u\right)+u+C$$
Or, putting in $\ln x$,
$$\ln\left(\ln\left(bx\right)\right)\left(\ln a-\ln b\right)+\ln x+C$$
But according the Wolfram Alpha I'm missing the $x$ value inside the parentheses:
$$\log(x)+(\log(a x)-\log(b x)) \log(\log(b x))+C$$
And I'm not sure where I dropped them...

Notice, follow the simple steps $$\int\frac{\ln a+u}{\ln b+u}\ du=\int \frac{(\ln b+u)+(\ln a-\ln b)}{\ln b+u}\ du $$ $$=\int du+(\ln a-\ln b)\int \frac{1}{\ln b+u}\ du $$
$$=u+(\ln a-\ln b)\ln(\ln b+u)+C $$ Now, setting $u=\ln x\ \ (\forall \ \ \color{red}{a, b, x>0})$ $$=\ln x+(\ln a-\ln b)\ln(\ln b+\ln x)+C$$
 $$=\ln x+((\ln a+\ln x)-(\ln b+\ln x))\ln(\ln (bx))+C$$
$$=\ln x+(\ln(ax)-\ln(bx))\ln(\ln(bx))+C$$
