求下面这个不定积分,感激!
解：∫[(xarctanx)/(1+x²)^(5/2)]dx
=(1/2)∫[(arctanx)/(1+x²)^(5/2)]dx²
=(1/2)∫[(arctanx)(1+x²)^(-5/2)]d(x²+1)
=(1/2)(-2/3)∫arctanxd(x²+1)^(-3/2)
=(-1/3)arctanx(x²+1)^(-3/2)-∫[(x²+1)^(-3/2)]darctanx
后一个积分，令arctanx=u，则x=tanu，sinu=x/√(1+x²)，故
∫[(x²+1)^(-3/...全部
解：∫[(xarctanx)/(1+x²)^(5/2)]dx
=(1/2)∫[(arctanx)/(1+x²)^(5/2)]dx²
=(1/2)∫[(arctanx)(1+x²)^(-5/2)]d(x²+1)
=(1/2)(-2/3)∫arctanxd(x²+1)^(-3/2)
=(-1/3)arctanx(x²+1)^(-3/2)-∫[(x²+1)^(-3/2)]darctanx
后一个积分，令arctanx=u，则x=tanu，sinu=x/√(1+x²)，故
∫[(x²+1)^(-3/2)]darctanx
=∫[(tan²u+1)^(-3/2)]du
=∫cos³udu
=∫cos²udsinu
=∫(1-sin²u)dsinu
=sinu-(1/3)sin³u+C
=x/√(1+x²)-(1/3)[x/√(1+x²)]³+C
故原积分为(-1/3)arctanx(x²+1)^(-3/2)-x/√(1+x²)+(1/3)[x/√(1+x²)]³+C
=(-1/3)arctanx(x²+1)^(-3/2)-x(3+2x²)/[3(1+x²)^(3/2)]+C。
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