Question

1/(1+t^2)^2的积分是什么

Answer 1

∫[1/(1+t²)²]dt,令t=tanu,dt=sec²udu

=∫[sec²u/(1+tan²u)²]du

=∫(sec²u/sec^4u)du

=∫(1/sec²u)du

=∫cos²udu

=(1/2)∫(1+cos2u)du

=(1/2)∫du+(1/2)(1/2)∫cos2ud(2u)

=(1/2)u+(1/4)sin2u+C

=(1/2)u+(1/4)*2sinucosu+C

=(1/2)arctant+(1/2)[t/√(1+t²)][1/√(1+t²)]+C

=(1/2)arctant+(1/2)[t/(1+t²)]+C

=(1/2)arctant+t/[2(1+t²)]+C