Question

∫cosx/[sinx(1+sinx)^2]dx=?求解答过程

Answer 1

令t=sinx cosxdx=dt

原式=∫dt/t(1+t)^2

=∫(1+t-t)/t(1+t)^2dt

=∫[1/t(1+t)-1/(1+t)^2]dt

=∫[1/t-1/(1+t)-1/(1+t)^2]dt

=ln|t|-ln|1+t|+1/(1+t)+C

=ln|sinx/(1+sinx)|+1/(1+sinx)+C