Question

高数 积分求详解

Answer 1

let
u=1+e^x
du = e^x dx
x=-1, u=1+e^(-1)
x=0, u=2
∫(-1->1) f(x) dx

=∫(-1->0) f(x) dx + ∫(0->1) f(x) dx
=∫(-1->0) dx/(1+e^x) + ∫(0->1) dx/(1+x)
=∫(-1->0) dx/(1+e^x) + [ln(1+x)]|(0->1)
=∫(1+e^(-1)->2) [du/(u-1)]/u + ln2
=ln2 +∫(1+e^(-1)->2) [1/(u-1) -1/u] du
=ln2 + [ln(u-1)/u]|(1+e^(-1)->2)
=ln2 + ln(1/2) - ln[ e^(-1)/(1+e^(-1)) ]
= -ln [1/(1+e) ]
=ln(1+e)

Answer 2