兩題分式求不定績分20點

1.∫dx/(x^6-1)

=(1/2) ∫[1/(x^3-1)-1/(x^3+1)] dx~接下來就用分式分解後再做

=(1/2){[(-1/6)ln(x^2+x+1)+(1/3)ln(x-1)-(tan^-1[(2x+1)/3]/3)]-[(-1/6)ln(x^2-x+1)+(1/3)ln(x+1)+(tan^-1[(2x-1)/3]/3)]+C

2.∫dy/[y(2y^3+1)^2]

=∫y^2/[y^3(2y^3+1)^2] dy, let u=(2y^3+1), du=6y^2dy

=∫(1/6)/{[(u-1)/2]u^2} du

=(1/3)∫ 1/[u^2(u-1)] du

=(1/3)∫[1/(u-1)-1/u-1/u^2] du

=(1/3)[ln(u-1)-ln(u)+1/u]

=(1/3)ln(2y^3)-ln(2y^3+1)+1/(2y^3+1)+C

2010-05-20 18:38:59 補充：

1.忘了補上根號

=(1/2){[(-1/6)ln(x^2+x+1)+(1/3)ln(x-1)-(tan^-1[(2x+1)/√3]/√3)]-[(-1/6)ln(x^2-x+1)+(1/3)ln(x+1)+(tan^-1[(2x-1)/√3]/√3)]+C

另外 怕你看不懂

1/(x^3-1)=[(-1/3)/(x-1)+(x/3+2/3)/(x^2+x+1)

=(-1/3)/(x-1)+(x/3+1/6)/(x^2+x+1)+(1/2)/(x^2+x+1)