求不定积分x*lnx⼀(1+x^2)^(3⼀2)！过程也写一下谢谢啦！！！

解：∫x*lnx/(1+x^2)^(3/2)dx
=1/2*∫lnx*d(1+x^2)/(1+x^2)^(3/2)dx
=1/2*∫芦迅枣lnx*d[-2/(1+x^2)^(1/2)]
=-∫lnx*d[1/(1+x^2)^(1/2)]
=∫[1/(1+x^2)^(1/2)]*1/x*dx-(lnx)/[(1+x^2)^(1/2)]
积分式中令x=tant,则dx=sec^2tdt
=∫1/sect*cott*sec^2tdt-(lnx)/[(1+x^2)^(1/2)]
=∫csctdt-(lnx)/[(1+x^2)^(1/2)]
=ln|csct-cott|-(lnx)/[(1+x^2)^(1/2)]+C
=ln|√(1+x^2)/x-1/x|-(lnx)/[(1+x^2)^(1/2)]+C
其中：
∫cscx dx
=∫1/sinx dx
=∫1/[2sin(x/2)cos(x/2)] dx
=∫1/[sin(x/2)cos(x/陪拆昌磨2)] d(x/2)
=∫1/tan(x/2)*sec²(x/2) d(x/2)
=∫1/tan(x/2) d[tan(x/2)]
=ln|tan(x/2)|+C
=ln|sin(x/2)/cos(x/2)|+C
=ln|2sin(x/2)cos(x/2)/[2cos²(x/2)]|+C
=ln|sinx/(1+cosx)|+C
=ln|sinx(1-cosx)/sin²x|+C
=ln|(1-cosx)/sinx|+C
=ln|cscx-cotx|+C