Can someone show me if my steps are wrong in solving $\int \frac{\sqrt{1+\ln x}}{x \ln x} dx$

Question

I just need to know if this method of solving this integral $$ \int \frac{\sqrt{1+\ln x}}{x \ln x} dx $$ is correct or not and if not where am I wrong?

Answer 1

You may just perform the change of variable \begin{equation*} v=\sqrt{1+\ln x},\quad \ln x=v^2-1,\quad x=e^{v^2-1}, \quad dx=2ve^{v^2-1}dv,\quad \frac{dx}x =2v\:dv \end{equation*} to get $$\begin{align} \int \frac{\sqrt{1+\ln x}}{x \ln x} dx&=\int \frac{v}{v^2-1} 2v\:dv\\\\ &=2\int \frac{v^2}{v^2-1}dv\\\\ &=2\int dv+2\int \frac{1}{v^2-1}dv\\\\ &=2v+\log \left|\frac{1-v}{1+v}\right|+C\\\\ &=2\sqrt{1+\ln x}+ \log\left|\frac{\sqrt{1+\ln x}-1}{\sqrt{1+\ln x}+1}\right|+C. \end{align} $$ Thus what you did is correct.