let
x=lny
dx= 1/y dy
∫f(x)dx
= ∫(1/y)f(lny)dy
=∫ln(y+1)/y^2 dy
=-∫ln(y+1)d(1/y)
= -ln(y+1)/y +∫ 1/(y(y+1)] dy
=-ln(y+1)/y +∫ [1/y-1/(y+1)] dy
= -ln(y+1)/y + ln|y/(y+1)| + C
=-[ln(e^x+1)]/e^x + ln| e^x/(e^x+1)| + C