Question

几道求不定积分的题,求过程

1, ∫ 1/( (x-2)^2*(x-3) ) dx

2, ∫ sinx * sin2x * sin3x dx

3, ∫(x^2 - 5x + 9) / ( x^2 - 5x +6 ) dx

4, ∫ cosx / ( sinx( 1 + sinx)^2 ) dx

答案:
1, 1/(x-2) + In| (x-3)/(x-2)| + c
2, 1/8*(1/3 * cos(6x) - 1/2 * cos(4x) - cos(2x)) +c
3, x+3In|(x-3)/(x-2)| +c
4, In|sinx/(1+sinx)| + 1/(1+sinx) +c

求详细过程,解题思路..........................................

Answer 1

1、
设1/[(x-2)²(x-3)] = A/(x-2)²+B/(x-2)+C/(x-3)
解得：A=-1，B=-1，C=1
∫dx/[(x-2)²(x-3)]
=-∫dx/(x-2)²-∫dx/(x-2)+∫dx/(x-3)，有理积分法
=1/(x-2)-ln|x-2|+ln|x-3| + C
=1/(x-2)+ln|(x-3)/(x-2)| + C
2、
∫(sinx)(sin2x)(sin3x) dx
=(1/2)∫(sin3x)(cosx-cos3x) dx，前两项，积化和差
=(1/2)∫sin3xcosx dx - (1/2)∫sin3xcos3x dx
=(1/4)∫(sin2x+sin4x) - (1/6)∫sin3xcos3x d(3x)，前面的部分，积化和差
=(1/8)∫sin2x d(2x) + (1/16)∫sin4x d(4x) + (1/6)∫cos3x d(cos3x)
=(-1/8)cos2x - (1/16)cos4x + (1/12)cos²(3x) + C
=(-1/8)cos2x - (1/16)cos4x + (1/24)cos6x + C'
3、
(x²-5x+9)/(x²-5x+6)=(x²-5x+6+3)/(x²-5x+6)=1+3/[(x-2)(x-3)]
设3/[(x-2)(x-3)]=A/(x-2)+B/(x-3)
解得：A=-3，B=3
∫(x²-5x+9)/(x²-5x+6) dx
=∫ dx - 3∫dx/(x-2) + 3∫dx/(x-3)，有理积分法
=x - 3ln|x-2| + 3ln|x-3| + C
=x + 3ln|(x-3)/(x-2)| + C
4、
令cosx/[sinx(1+sinx)²]=A/sinx + B/(1+sinx) + C/(1+sinx)²
A=cosx，B=-cosx，C=-cosx
∫cosx/[sinx(1+sinx)²] dx
=∫cosx/sinx dx - ∫cosx/(1+sinx) dx - ∫cosx/(1+sinx)² dx，有理积分法
=∫d(sinx)/sinx - ∫d(sinx+1)/(1+sinx) - ∫d(sinx+1)/(1+sinx)²
=ln|sinx| - ln|1+sinx| -[-1/(1+sinx)²] + C
=ln|sinx/(1+sinx| + 1/(1+sinx)² + C