反常积分问题 高等数学反常积分的一个问题
\u53cd\u5e38\u79ef\u5206\u95ee\u9898\u62c6\u9879 1+x^4=(1+x2)2-2x2=(1+\u221a2x+x2)(1-\u221a2x+x2) =\u222b1/2\u221a2x(1/(1-\u221a2x+x2)-1/(1+\u221a2x+x2))dx =1/2\u221a2\u222b1/x-(x-\u221a2)/(1-\u221a2x+x2)-1/x+(x+\u221a2)/(1+\u221a2x+x2)dx =1/2\u221a2\u222b(x+\u221a2)/(1+\u221a2x+x2)-(x-\u221a2)/(1-\u221a2x+x2)dx \u7136\u540e\u7ee7\u7eed\u62c6\u9879\uff0c 1+\u221a2x+x2=(x+\u221a2/2)2+1/2 \u524d\u8005\u4e09\u89d2\u6362\u5143x=tanu/2-\u221a2/2\uff0c\u540e\u8005\u4e09\u89d2\u6362\u5143x=tanu/2+\u221a2/2\u5373\u53ef\u8ba1\u7b97\u4e0d\u5b9a\u79ef\u5206
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应该是这样吧？不对的话再交流
(4) I = ∫<0, +∞> dx/[(1+x)(1+x^2)]
= (1/2)∫<0, +∞> [1/(1+x) + (1-x)/(1+x^2)]dx
= (1/2)∫<0, +∞> [1/(1+x) + 1/(1+x^2) - x/(1+x^2)]dx
= (1/2)[ln(1+x) + arctanx - (1/2)ln(1+x^2)]<0, +∞>
= (1/2) [ln{(1+x)/√(1+x^2)} + arctanx]<0, +∞>
= (1/2)[lim<x→+∞>ln{(1+x)/√(1+x^2)} + π/2]
= (1/2)[lim<x→+∞>ln{(1/x+1)/√(1/x^2+1)} + π/2] = π/4
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