Integral Question

let x=tanθ
dx/dθ=sec^2θ
∫1 / (x^2 +1) ^2 dx
=∫sec^2θ / (tan^2θ +1) ^2 dθ
=∫sec^2θ / sec^4θ dθ
=∫1 / sec^2θ dθ
=∫cos^2θ dθ
=∫1/2 + 1/2 cos(2θ ) dθ
=(1/2)θ +1/4 sin(2θ)+C

2007-01-10 15:35:22 補充：
轉番做x就是(1/2)(arctan x) 1/4 sin(2arctan x) C

2007-01-10 15:45:47 補充：
可以更簡considerx=tanθtan(2θ)= 2tanθ/(1−tan^2θ)=2x/(1-x^2)sin(2θ)=x/(1+x^2)so∫1 / (x^2 +1) ^2 dx=(1/2)(arctan x)+ 1/2 [x/(1+x^2)] C