求 t^2/(t^4-1)dt 的积分
1个回答

∫[t^2/(t^4-1)]dt

=(1/2)∫[(t^2+1+t^2-1)/(t^4-1)]dt

=(1/2)∫[1/(t^2-1)]dt+(1/2)∫[1/(t^2+1)]dt

=(1/4)∫[(t+1-t+1)/(t^2-1)]dt+(1/2)arctant

=(1/4)∫[1/(t-1)]dt-(1/4)∫[1/(t+1)]dt+(1/2)arctant

=(1/4)ln|t-1|-(1/4)ln|t+1|+(1/2)arctant+C