Question

求解答：用第一类换元法求不定积分。

Answer 1

1)、
令u = √x + 1，x = (u - 1)² and dx = 2(u - 1) du
∫ cos(√x + 1)/√x dx
= ∫ 2(u - 1)cos(u)/(u - 1) du
= 2∫ cos(u) du
= 2sin(u) + C
= 2sin(√x + 1) + C
2)、
∫ dx/(4x² - 1)
= ∫ dx/[(2x + 1)(2x - 1)]
= (1/2)∫ [(2x + 1) - (2x - 1)]/[(2x + 1)(2x - 1)] dx
= (1/2)∫ [1/(2x - 1) - 1/(2x + 1)] dx
令u = 2x - 1 and v = 2x + 1，then du = 2dx and dv = 2dx
= (1/4)∫ du/u - (1/4)∫ dv/v
= (1/4)ln|u| - (1/4)ln|v| + C
= (1/4)ln|u/v| + C
= (1/4)ln|(2x - 1)/(2x + 1)| + C