高数,用换元积分法求积分 ∫1/(e^x-e^-x)dx

∫1/(e^x-e^-x)dx
= ∫e^x/(e^2x-1)dx
令t=e^x x=lnt dx=1/tdt
原式=∫[t/(t^2-1)]*(1/t)dt
=∫[1/(t^2-1)]dt
=(1/2)∫{[1/(t-1)]-[1/(t+1)]}dt
=(1/2){∫[1/(t-1)]d(t-1)-∫[1/(t+1)]}d(t+1)}
=(1/2){ln|t-1|-ln|t+1|}+c
=ln{[(t-1)/(t+1)]^(1/2)}+c
=ln{[(e^x-1)/(e^x+1)]^(1/2)}+c