Question

求不定积分∫x^2/(1+x^2)^2 dx

Answer 1

设 x=tant,dx=(sect)^2dt,
t=arctanx,1+x^2=(sect)^2,cost=1/√（1+x^2),
sint=x/√（1+x^2),
sin2t=2sintcost=2x/(1+x^2)
原式=∫(tant)^2(sect)^2dt/*(sect)^4
=∫(sint)^2*(cost)^2dt/(cost)^2
=∫（sint)^2dt
=(1/2)∫（1-cos2t)dt
=t/2-(1/4)sin2t+C
=(1/2)arctanx-x/[2(1+x^2)]+C.

Answer 2

x=tanu, dx=sec^2udu
∫x^2/(1+x^2)^2 dx
=∫tan^2u * sec^2u / sec^4u du
=∫sin^2udu
=1/2∫(1-cos2u)du
=1/2u-1/4sin2u+C
=1/2u-1/2tanu/(1+tan^2u)+C
=1/2arctanx-x/[2(1+x^2)]+C