For `x>0`, let `f(x)=int_1^x log_et/(1+t)dt` find the function `f(x)+f(1/x)` and show that `f(e)+f(1/e)=1/2`

For `x>0`, let `f(x)=int_1^x log_et/(1+t)dt` find the function `f(x)+f

1.	For `x>0`, let `f(x)=int_1^x log_et/(1+t)dt` find the function `f(x)+f(1/x)` and show that `f(e)+f(1/e)=1/2`
Answer» `f(x)=int_1^x (log_e^t)/(t+1)dt-(1)` `f(1/x)=int_1^x(log_e^t)/(t+1)dt` `Let t=1/h,dt=-1/h^2dt` `if t=1,h=1 or t=1/x,4=x` `f(1/x)=int_1^x(log_e(1/h))/(1+1/h)(-1/h^2dx)` `f(1/x)=int_1^x(-logh(-1))/(h+1)h dh` `f(1/x)=int_1^x9(log_e^t)/(t(t+1))dt-(2)` Adding 1 and 2 `f(x)+f(1/x)=int_1^x(log_e^t)/(t+1)(1+1/t)dt` `int_1^x(log_e^t)/(t+1)(1+1/t)dt` `int_1^x(log_e^t)/t dt` `logt=v` `1/tdt=dv` `f(x)+f(1/x)=int_0^(logx)vdv` `=[v^2/2]_0^logx` `=(log_e^x)^2/2` `f(e)+f(1/e)=(log_e^e)^2/2=1/2`