1、∫3⼀x^3+1dx 2、∫dx⼀x（x^6+4）

1、∫ 3/(x³+1) dx
=∫ 3/[(x+1)(x²-x+1)] dx
令3/[(x+1)(x²-x+1)]=A/(x+1)+(Bx+C)/(x²-x+1)
右边通分相加与左边比较系数，得：A=1，B=-1，C=2
=∫ 1/(x+1) dx - ∫ (x-2)/(x²-x+1) dx
=ln|x+1| - (1/2)∫ (2x-1-3)/(x²-x+1) dx
=ln|x+1| - (1/2)∫ (2x-1)/(x²-x+1) dx + (1/2)∫ 3/(x²-x+1) dx
=ln|x+1| - (1/2)∫ 1/(x²-x+1) d(x²-x) + (3/2)∫ 1/[(x-1/2)²+3/4] dx
=ln|x+1| - (1/2)ln(x²-x+1) + √3arctan[(2x-1)/√3] + C

2、∫ 1/[x(x^6+4)] dx
分子分母同乘以x^5
=∫ x^5/[x^6(x^6+4)] dx
=(1/6)∫ 1/[x^6(x^6+4)] d(x^6)
=(1/24)∫ [1/x^6 - 1/(x^6+4)] d(x^6)
=(1/24)[lnx^6 - ln(x^6+4)] + C
=(1/4)ln|x| - (1/24)ln(x^6+4) + C

【数学之美】团队为您解答，若有不懂请追问，如果解决问题请点下面的“选为满意答案”。