求∫[xlnx/(1+x2)3/2]dx
∫[xlnx/(1+x2)3/2]dx =1/2∫lnx•(1+x2)-(3/2)d(1+x2) =∫lnxd[-1/√(1+x2)] =-[lnx/√(1+x2)]+∫(1/√(1+x2)d(lnx) =-lnx/√(1+x2)+∫[1/x√(1+x2)]dx (令x=tant) =-lnx/√(1+x2)+∫[(1/(tant•sect)]d(tant) =-lnx/√(1+x2)+∫csctdt =-lnx/√(1+x2)+ln∣csct-cott∣+C =-lnx/√(1+x2)+ln∣√(1+2)/x-1/x∣+C
