求∫[xlnx/(1+x²)^3/2]dx

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