Question

不定积分∫x^2/(x^2+1)^2dx怎么求？

Answer 1

令x=tanu，则x²+1=sec²u，dx=sec²udu
∫x^2/(x^2+1)^2dx

=∫ [tan²u/(secu)^4]sec²udu
=∫ tan²u/sec²udu
=∫ (sec²u-1)/sec²udu
=∫ 1 du - ∫ cos²u du
=u - (1/2)∫ (1+cos2u) du
=u - (1/2)u - (1/4)sin2u + C
=(1/2)u - (1/2)sinucosu + C
=(1/2)arctanx - (1/2)x/(1+x²) + C

如果解决了问题，请采纳

Answer 2

∫x^2/(x^2+1)^2dx

=∫(x^2+1-1)/(x^2+1)^2dx
=∫1/(x²+1)dx-∫1/(x²+1)²dx
=arctanx-这个有个递推公式的自己套吧。