Question

(3x-2)/(x^2-2x+10)不定积分，要详细过程

Answer 1

答：(3/2)ln(x²-2x+10) + (1/3)arctan[ (x-1)/3 ] + C

d/dx (x²-2x+10) = 2x-2，凑微分
3x-2 = 3*((2x-2)+2)/2-2 = (3/2)(2x-2)+1
∫ (3x-2)/(x²-2x+10) dx
= (3/2)∫ (2x-2)/(x²-2x+10) dx + ∫ dx/(x²-2x+10)
= (3/2)∫ d(x²-2x+10)/(x²-2x+10) + ∫ d(x-1)/[(x-1)²+3²]
= (3/2)ln(x²-2x+10) + (1/3)arctan[ (x-1)/3 ] + C

追问

∫ d(x-1)/[(x-1)²+3²]变到下一步 (1/3)arctan[ (x-1)/3 ]
我觉得是(1/9)arctan[ (x-1)/3 ] 为什么是1/3呢

追答

公式∫ dx/(x²+a²) = (1/a)arctan(x/a)

令x = atanz,dx = asec²zdz
1/(x²+a²) dx = 1/(a²sec²z)*asec²z dz
= z/a + C
= (1/a)arctan(x/a) + C

追问

但是(1/3)arctan[ (x-1)/3 ]算出来是(1/3)*1/[1+(x-1)^2/9)]=3/[(x-1)²+3²]，原式却是1/[(x-1)²+3²]啊

追答

d/dx (1/3)arctan[ (x-1)/3 ]
= 1/3 * 1/[1+((x-1)/3)²] * 1/3
= 1/9 * 1/[1+(x-1)²/9]
= 1/9 * 9/[9+(x-1)²]
= 1/[9+(x²-2x+1)]
= 1/(x²-2x+10)

追问

1/3 * 1/[1+((x-1)/3)²] * 1/3最后* 1/3是(x-1)/3的系数是吗？就相当于对(x-1)/3求导？

追答

复合函数求导
除了要对arctan求导外，还要对里面的(x-1)/3再求导一次

y = (1/3)arctan[ (x-1)/3 ]
由链式法则，
dy/dx = (1/3) d(arctanu)/du * du/dx
这里u = (x-1)/3