帮忙解个不定积分∫1/(x^3+a)dx

Question

帮忙解个不定积分
∫1/(x^3 + a) dx

Answer 1

很高兴为您解答.
我们先计算∫ 1/(x³ + a³) dx,然后在答案中再替代回a = a^(1/3)即可
∫ 1/(x³ + a³) dx
= ∫ 1/[(x + a)(x² + ax + a²)] dx
= (1/(3a²))∫ dx/(a + x) - (1/(3a²))∫ (x - 2a)/(x² - ax + a²) dx
= (1/(3a²))ln(a + x) - (1/(3a²))∫ [(2x - a)/2 - 3a/2]/(x² - ax + a²) dx
= (1/(3a²))ln(a + x) - (1/(6a²))∫ d(x² - ax + a²)/(x² - ax + a²) + (1/(2a))∫ d(x - a/2)/[(x - a/2)² + 3a²/4]
= (1/(3a²))ln(a + x) - (1/(6a²))ln(x² - ax + a²) + (1/(2a))(2/(√3a))arctan[(x - a/2)(2/(√3a))] + C
= (1/(3a²))ln(a + x) - (1/(6a²))ln(x² - ax + a²) + (1/(√3a²))arctan[2x/(√3a) - 1/√3] + C
现在用a = a^(1/3)替换,a² = a^(2/3),可得：
∫ 1/(x³ + a) dx
= 1/(3a^(2/3)) * ln[a^(1/3) + x] - 1/(6a^(2/3)) * ln[x² - (a^(1/3))x + a^(2/3)]
+ 1/(√3a^(2/3)) * arctan[2x/(√3a^(1/3)) - 1/√3] + C