Question

无理函数积分

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Answer 1

作变换：t=(2+x)^(1/3)
则x=t^3-2
dx=3t^2dt
原积分化为有理函数的积分了：
原式=∫(t^3-2)t*3t^2/(t^3-2+t)dt
=3∫(t^6-2t^3)/(t^3+t-2)dt
=3∫[t^6+t^4-2t^3-t^4-t^2+2t+t^2-2t]/(t^3+t-2)dt
=3∫[t^3-t+(t^2-2t)/(t^3+t-2)]dt
=3∫[t^3-t+(t^2-2t)/(t^3-t+2t-2)]dt
=3∫[t^3-t+(t^2-2t)/(t-1)(t^2+t+2)]dt
=3∫[t^3-t+a/(t-1)+(bt+c)/(t^2+t+2)]dt
其中待定系数：a(t^2+t+2)+(bt+c)(t-1)=t^2-2t
(a+b)t^2+t(a-b+c)+2a-c=t^2-2t
a+b=1, a-b+c=-2,2a-c=0
解得：a=-0.25, b=1.25,c=-0.5
这样上式=3∫[t^3-t-0.25/(t-1)+(1.25t-0.5)/(t^2+t+2)]dt
这样就可以直接积分了。