上限是1,下限是0的x的平方/(1+x的平方)的平方的定积分

Question

上限是1,下限是0的x的平方/(1+x的平方)的平方的定积分

Answer 1

令x = tanθ,dx = sec²θ dθ
∫(0,1) x²/(1 + x²)² dx
= ∫(0,π/4) [tan²θ/(1 + tan²θ)²](sec²θ) dθ
= ∫(0,π/4) (tan²θsec²θ)/(sec²θ)² dθ
= ∫(0,π/4) (sin²θ/cos²θ)(cos²θ) dθ
= ∫(0,π/4) sin²θ dθ
= ∫(0,π/4) (1 - cos2θ)/2 dθ
= [θ/2 - (1/4)sin2θ] |(0,π/4)
= (π - 2)/8