Question

求1/(x的三次方-1)的不定积分

Answer 1

d(x+1/2)

=1/3*ln(x-1)-1/6*ln(x²+x+1)-√(4/3)*arctan[√(4/3)*(x+1/2)]+C

=1/3*ln(x-1)-1/6*ln(x²+x+1)-2/√3*arctan[(2x+1)/√3)]+C

=(1/3)ln(x-1)-(1/6)ln(x²+x+1)-(2/√3)arctan[(2x+1)/√3]+C

不定积分的公式：

1、∫ a dx = ax + C，a和C都是常数

2、∫ x^a dx = [x^(a + 1)]/(a + 1) + C，其中a为常数且 a ≠ -1

3、∫ 1/x dx = ln|x| + C

4、∫ a^x dx = (1/lna)a^x + C，其中a > 0 且 a ≠ 1

5、∫ e^x dx = e^x + C

6、∫ cosx dx = sinx + C

7、∫ sinx dx = - cosx + C

8、∫ cotx dx = ln|sinx| + C = - ln|cscx| + C

Answer 2

这题目很烦的:答案是(1/3)ln(x-1)-(1/6)ln(x²+x+1)-(2/√3)arctan[(2x+1)/√3]+C
过程:
∫1/(x³-1)
dx
=∫1/(x-1)(x²+x+1)
dx
=∫1/3(x-1)-(x+2)/3(x²+x+1)]
dx,总之在/后面的都是分母
=1/3*∫1/(x-1)
dx-1/3*∫(x+2)/(x²+x+1)
dx
=1/3*∫1/(x-1)
d(x-1)-1/3*∫[(2x+1)/2(x²+x+1)+3/2(x²+x+1)]
dx
=1/3*ln(x-1)-1/6*∫(2x+1)/(x²+x+1)
dx-1/2*∫1/(x²+x+1)
dx
=1/3*ln(x-1)-1/6*∫(2x+1)d(x²+x+1)/(x²+x+1)(2x+1)-1/2*∫1/[(x+1/2)²+3/4]
dx
=1/3*ln(x-1)-1/6*ln(x²+x+1)-1/2*∫1/[(x+1/2)²+3/4]
d(x+1/2)
=1/3*ln(x-1)-1/6*ln(x²+x+1)-√(4/3)*arctan[√(4/3)*(x+1/2)]+C
=1/3*ln(x-1)-1/6*ln(x²+x+1)-2/√3*arctan[(2x+1)/√3)]+C
=(1/3)ln(x-1)-(1/6)ln(x²+x+1)-(2/√3)arctan[(2x+1)/√3]+C