问题
解答题
求不定积分∫
|
答案
令1+ex=t，
则dt=exdx=（t-1）dx，dx=
.dt t-1
∴∫
=∫dx (1+ex)2 dt (t-1)t2
=∫(
-1 t(t-1)
)dt1 t2
=∫(
-1 t-1
-1 t
)dt1 t2
=ln(t-1)-lnt+
+C1 t
=lnex-ln(1+ex)+
+C1 1+ex
=x-ln(1+ex)+
+C．1 1+ex
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