How to find $\int \frac {\sinh(\ln x)} {2x}\,dx$?

Question

I've tried

$$\int \frac {\sinh(\ln x)} {2x}\,dx = \int \frac {e^{\ln x}-e^{-\ln x}} {2x}\,dx = \int \frac {x-e^{-\ln x}} {2x}\,dx$$

Answer 1

prove that $$\sinh(\ln(x))=\frac{1}{2}x-\frac{1}{2x}$$

Answer 2

If you set $t=\ln x$, then $dt=\frac{1}{x}\,dx$ and the integral becomes $$ \int \sinh t\,dt = \cosh t+c=\cosh \ln x+c $$ Then you can transform this as you please: $$ \cosh\ln x=\frac{e^{\ln x}+e^{-\ln x}}{2}=\frac{x+1/x}{2}=\frac{x^2+1}{2x} $$

With your original attempt, you just have to recognize that $e^{-\ln x}=\dfrac{1}{x}$