網友廻答：
- ∫sinxdx/(sinx^3 cosx^3)
=∫dx/sinx^2(1 cotx^3)
=-∫dcotx/(1 cotx^3)
cotx=u
=-∫du/(1 u^3)
=(-1/6)ln|u^2-u 1| (1/√3)arctan[(2u-1)/√3] (1/3)ln|u 1| C
=(-1/6)ln|cotx^2-cotx 1| (1/√3)arctan[(2cotx-1)/√3] (1/√3ln|cotx 1| C
∫dx/(1 x^3)=∫dx/[(1 x)(1 x^2-x)]=(1/3)∫(x 1)^2-(x^2-x 1)dx/[x(1 x)(1-x x^2)]
=(1/3)∫(x 1)dx/x(1-x x^2)-(1/3)∫dx/x(1 x)
=(1/3)∫dx/x(1-x x^2) (1/3)∫dx/(1-x x^2) (1/3)ln(1 x)/x
=(1/6)∫dx^2/x^2(1-x x^2) ...
=(1/6)∫(x-1)dx^2/[x^2(1-x x^2)(x-1)] ..
=(1/6)∫dx^2/(x-1)(1-x x^2)-(1/6)∫dx^2/x^2(x-1) ...
=(1/3)∫dx/(x-1)(1-x x^2)-(1/3)∫dx/x(x-1) ...
=(1/3)∫dx/(x-1)-(1/3)∫(x-1)dx/(1-x x^2)-(1/3)ln(x-1)/x ...
=(1/3)ln(x-1)-(1/6)∫d(x^2-x 1)/(x^2-x 1) (1/6)∫dx/(1-x x^2)-(1/3)ln(x-1)/x ...
=(-1/6)ln|x^2-x 1| (1/√3)arctan[(2x-1)/√3] (1/3)ln|x 1| C
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