对1／（4x^2）（1+4x^2）微积分,

Question

对1／（4x^2）（1+4x^2）微积分,

Answer 1

微分：
d/dx 1/[(4x²)(1 + 4x²)] = d/dx 1/(16x⁴ + 4x²)
= - (16x⁴ + 4x²)'/(16x⁴ + 4x²)²
= - (16 * 4x³ + 4 * 2x)/[(4x²)²(1 + 4x²)²]
= - 8x(8x² + 1)/(16x⁴(1 + 4x²)²]
= - (1 + 8x²)/[2x³(1 + 4x²)²]
积分：
∫ 1[(4x²)(1 + 4x²)] dx
= ∫ [(4x² + 1) - 4x²]/[(4x²)(1 + 4x²)] dx
= ∫ [1/(4x²) - 1/(1 + 4x²)] dx
= (1/4)∫ 1/x² dx - (1/2)∫ 1/[1 + (2x)²] d(2x)
= (1/4)(- 1/x) - (1/2)arctan(2x) + C
= - 1/(4x) - (1/2)arctan(2x) + C