∫1/x^2(x+2)dx.∫x^3/(3+x)dx.∫1/(x-2)^2(x-3)dx.

Question

∫1/x^2(x+2)dx.∫x^3/(3+x) dx.∫1/(x-2)^2(x-3)dx.

Answer 1

1、∫ 1/[x²(x+2)] dx
=(1/2)∫ 2/[x²(x+2)] dx
=(1/2)∫ (2+x-x)/[x²(x+2)] dx
=(1/2)∫ (2+x)/[x²(x+2)] dx - (1/2)∫ x/[x²(x+2)] dx
=(1/2)∫ 1/x² dx - (1/2)∫ 1/[x(x+2)] dx
=(1/2)∫ 1/x² dx - (1/4)∫ [1/x-1/(x+2)] dx
=-1/(2x) - (1/4)ln|x| + (1/4)ln|x+2| + C
2、∫ x³/(3+x) dx
=∫ (x³+27-27)/(3+x) dx
=∫ (x³+27)/(3+x) dx - 27∫ 1/(3+x) dx
=∫ (x²-3x+9) dx - 27∫ 1/(3+x) dx
=(1/3)x³ - (3/2)x² + 9x - 27ln|x+3| + C
3、∫ 1/[(x-2)²(x-3)] dx
=∫ (3-x+x-2)/[(x-2)²(x-3)] dx
=∫ (3-x)/[(x-2)²(x-3)] dx + ∫ (x-2)/[(x-2)²(x-3)] dx
=-∫ 1/(x-2)² dx + ∫ 1/(x-2)(x-3) dx
=-∫ 1/(x-2)² dx + ∫ [1/(x-3)-1/(x-2)] dx
=1/(x-2) + ln|x-3| - ln|x-2| + C
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