Question

求下面的不定积分

Answer 1

设 e^x + 1 = u，则 x = ln(u-1)。dx = du/(u-1)
那么，上面的不定积分变换为：
=∫du/(u-1) * 1/u²
=∫[1/(u-1) - 1/u - 1/u²]*du
=∫du/(u-1) - ∫du/u - ∫du/u²
=ln(u-1) - lnu + 1/u + C
=ln(e^x) - ln(e^x +1) + 1/(1+e^x) + C
=x - ln(e^x +1) + 1/(e^x + 1) + C

Answer 2

∫1/(1+e^x)² dx
=∫e^x/[e^x(1+e^x)²] dx
=∫1/[e^x(1+e^x)²] de^x
令e^x=u
=∫1/[u(1+u)²] du
=∫(1/u-1/(1+u)-1/(1+u)²] du
=ln|u|-ln|u+1|+1/(1+u)+C
=x-ln|e^x+1|+1/(e^x+1)+C