Question

麻烦帮我看下这道不定积分的题

麻烦帮我看下这道不定积分的题感恩大神

Answer 1

1/[(ax + b)ⁿ(cx + d)] = A/(ax + b)ⁿ + B/(ax + b)ⁿ⁻¹ + ...+ X/(ax + b)² + Y/(ax + b) + Z/(cx + d)
(ax + b)/[(cx² + dx + e)(fx + g)] = (Ax + B)/(cx² + dx + e) + C/(fx + g)
如果次方在括号外面的话,就根据第一种情况设
如果括号内是一个多项式的话,就根据第二种情况设,并且分子的次数要比分母少一次
如果括号内是多项式,而括号外又有高次方的话,就两种都用上
即1/[(ax² + bx + c)ⁿ(cx + d)]
= (Ax + B)/(ax² + bx + c)ⁿ + (Cx + D)/(ax² + bx + c)ⁿ⁻¹ + ...+ (Xx + Y)/(ax² + bx + c) + Z/(cx + d)
(x² + 1)/[(x + 1)²(x - 1)] = A/(x + 1) + B/(x + 1)² + C/(x - 1)
x² + 1 = A(x + 1)(x - 1) + B(x - 1) + C(x + 1)²
令x = - 1,2 = - 2B ==> B = - 1
令x = 1,2 = 4C ==> C = 1/2
令x = 0,1 = - A - (- 1) + 1/2 ==> A = 1/2
(x² + 1)/[(x + 1)²(x - 1)] = 1/[2(x + 1)] - 1/(x + 1)² + 1/[2(x - 1)]

∫ (x² + 1)/[(x + 1)²(x - 1)] dx
= (1/2)∫ dx/(x + 1) - ∫ dx/(x + 1)² + (1/2)∫ dx/(x - 1)
= (1/2)ln|x + 1| + 1/(x + 1) + (1/2)ln|x - 1| + C
= 1/(x + 1) + (1/2)ln|x² - 1| + C