e^(2x)/[e^x-e^(-x)]dx求不定积分.

Question

e^(2x)/[e^x-e^(-x)]dx求不定积分.

Answer 1

∫[e^(2x)/(e^x)-e(-x))]dx
=∫[e^(3x)/(e^(2x) -1)]dx
=∫[e^(2x)/(e^(2x) -1)]d(e^x)
=∫[e^(2x)-1+1]/[e^(2x)-1] d(e^(x)
=∫[1 +1/[e^(2x) -1] ] d(e^x)
=∫[1 +1/[(e^x+1)(e^x-1)] ] d(e^x)
=∫[1 +(1/2)[1/(e^x -1)- 1/(e^x +1)] ] d(e^x)
=e^x +(1/2)ln(e^x -1) -(1/2)ln(e^x +1) +C