简单积分题

令x = a sec t，则dx = a sec t tan t dt
代入得原式=∫a^4 (tan t)^4 sect dt
=a^4 ∫(sint)^4/(cost)^6 d(sint)
=a^4 ∫(sint)^4 d(sint)/[1-(sint)²]³
做代换u = sint，上式=a^4 ∫u^4 du/(1-u²)³
作有理分式分解u^4/(1-u²)³ = 1/(1-u²)³ - (1+u²)/(1-u²)² = 1/(1-u²)³+ 1/(1-u²) - 2/(1-u²)²
1/(1-u²) = 1/2 [1/(1-u) + 1/(1+u)]
2/(1-u²)² = 1/2 [1/(1-u)² + 1/(1+u)² + 1/(1+u) + 1/(1-u)]
1/(1-u²)³ = 1/8 [1/(1-u)³ + 1/(1-u) + 1/(1+u) + 1/(1-u)² + 1/(1+u)³ + 1/(1+u)² + 1/2 [1/(1-u)² + 1/(1+u)² + 1/(1+u) + 1/(1-u)]]
积出来把u换成sint，再代换成x就行了