Question

一道微积分题目

积[t^2/(1+t^2)^2]dt

Answer 1

解：
∫t^2/(1+t^2)^2 dt

=∫[1/(t^2+1)-1/(t^2+1)^2] dt

=∫1/(t^2+1) dt-∫1/(t^2+1)^2 dt 【令t=tanu，则dt=(secu)^2 du】

=∫1/(t^2+1) dt-∫(cosu)^2 du

=∫1/(t^2+1) dt-∫[1/2·cos2u+1/2]du

=∫1/(t^2+1) dt-1/2∫cos2u du-∫1/2du

=arctant-1/4·sin2u-u/2+C

=arctant-1/4·2t/(1+t^2)-(arctant)/2+C

=(arctant)/2-t/[2(1+t^2)]+C

Answer 2

没t=tanx
dt=[1/9conx)^2]dx
∫t^2/(1+t^2)^2 dt=∫(tanx)^2*(conx)^4/(conx^)2*dx
=∫(sinx)^2dx=∫(1-con2x)/2dx=∫1/2dx-∫con2xdx/2
=x/2-sin2x/4+c=arctanx/2-sin(2arctanx)/4+c
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