求1╱（1＋x∧2）∧2的不定积分

Answer 1

x = tanz，dx = sec²z dz
1/(1 + x²)² = 1/(1 + tan²z)² = 1/(sec²z)² = cos⁴z
原式= ∫ cos⁴z•sec²z dz
= ∫ cos²z dz
= (1/2)∫ (1 + cos2z) dz
= (1/2)[z + (1/2)sin2z] + C
= (1/2)arctan(x) + (1/2)[x/√(1 + x²)][1/√(1 + x²)] + C
= (1/2)arctan(x) + x/[2(1 + x²)] + C

Answer 2

1/2 (x/(1+x^2)+ArcTan[x])