integral calculus
posted by flo .
evaluate integral by substitution (e^xe^(x))/(e^x+e^(x))dx

Note that you have Integral(tanh(x) dx)
= Integral (sinh(x)/cosh(x) dx)
let u = cosh(x)
du = sinh(x) dx
and you have Integral(du/u) = ln(u) = ln(cosh(x)) = ln(e^x + e^x) + C
Equivalently, let u = e^x + e^x
Then du = e^x  e^x du
and you have du/u