Difficult Integral Involving the $\ln$ function

Question

Please help me solve this integral! I have tried multiple different procedures for integration by parts, as well as substitution and have not come up with anything. $$\int\frac{\ln x}{(\ln x+1)^2}dx$$

Thank you

Answer 1

Setting $\ln x=y\implies x=e^y$

$$\int\frac{\ln x}{(\ln x+1)^2}dx=\int\frac y{(y+1)^2}e^y dy$$

$$=\int\left(\frac{y+1-1}{(y+1)^2}\right) e^ydy=\int e^y\left(\frac1{y+1}-\frac1{(y+1)^2}\right)$$

If $\displaystyle f(y)=\frac1{y+1}, f'(y)=?$

Now, $$\int e^y\left[f(y)+f'(y)\right]dy=f(y)e^ydy+f'(y)e^y=d(e^yf(y))$$