零到正无穷广义积分(x^2+1)/(x^4-x^2+1)

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零到正无穷广义积分(x^2+1)/(x^4-x^2+1)零到正无穷广义积分(x^2+1)/(x^4-x^2+1)零到正无穷广义积分(x^2+1)/(x^4-x^2+1)如图所示∫(x^2+1)/(x^

零到正无穷广义积分(x^2+1)/(x^4-x^2+1)
零到正无穷广义积分(x^2+1)/(x^4-x^2+1)

零到正无穷广义积分(x^2+1)/(x^4-x^2+1)
如图所示

∫(x^2+1)/(x^4-x^2+1)dx = arctan[x/(1 - x^2)] + C
这个不定积分过程很麻烦,不写了,但是这个是绝对正确的,不信你可以求导试试
广义积分 ∫(x^2+1)/(x^4-x^2+1)dx ; x from 0 to infinite ,结果是 pi结果是pi吗?是不是0?像这类比较复杂的去凑导数公式有没有什么规律?。。。不可能是0 啊,你...

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∫(x^2+1)/(x^4-x^2+1)dx = arctan[x/(1 - x^2)] + C
这个不定积分过程很麻烦,不写了,但是这个是绝对正确的,不信你可以求导试试
广义积分 ∫(x^2+1)/(x^4-x^2+1)dx ; x from 0 to infinite ,结果是 pi

收起

∫(x^2+1)dx/(x^4-x^2+1)
=∫(x^2-1)dx/(x^4-x^2+1)+∫2dx/(x^4-x^2+1)
=(1/2)∫dx/(x^2-x-1) +(1/2)∫dx/(x^2+x-1)+∫dx/[x(x^2-x-1)]-∫dx/[x(x^2+x-1)]
=(1/2√5)ln[(x-1/2-√5/2)/(x-1/2+√5/2)]
+(1/2√5)...

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∫(x^2+1)dx/(x^4-x^2+1)
=∫(x^2-1)dx/(x^4-x^2+1)+∫2dx/(x^4-x^2+1)
=(1/2)∫dx/(x^2-x-1) +(1/2)∫dx/(x^2+x-1)+∫dx/[x(x^2-x-1)]-∫dx/[x(x^2+x-1)]
=(1/2√5)ln[(x-1/2-√5/2)/(x-1/2+√5/2)]
+(1/2√5)ln[(x+1/2-√5/2)/(x+1/2+√5/2)]
+(1/√5)∫dx/[x(x-1/2-√5/2)] -(1/√5)∫dx/[x(x-1/2+√5/2)
-(1/√5)∫dx/[x(x+1/2-√5/2) +(1/√5)∫dx/[x(x+1/2+√5/2)]
=(1/2√5)ln[(x-1/2-√5/2)/(x-1/2+√5/2)]
+(1/2√5)ln[(x+1/2-√5/2)/(x+1/2+√5/2)]
+(1/(√5/2+5/2))ln[(x-1/2-√5/2)/x] -(1/(√5/2-5/2))ln[(x-1/2+√5/2)/x]
-1/(√5/2-5/2)ln[x/(x+1/2-√5/2)+(1/(√5/2+5/2)ln(x/(x+1/2+√5/2)]
=(1/2√5)ln|[(x-1/2-√5/2)/(x-1/2+√5/2)]|
+(1/2√5)ln|[(x+1/2-√5/2)/(x+1/2+√5/2)]|+
(1/(√5/2+5/2))ln|[(x-1/2-√5/2)/(x+1/2+√5/2)]|
-1/(√5/2-5/2)ln|[(x-1/2+√5/2)/(x+1/2-√5/2)]|
∫[0,∞](x^2+1)dx/(x^4-x^2+1)
= -(1/2√5)ln[(1/2+√5/2)/(-1/2+√5/2)]
-(1/2√5)ln[(√5/2-1/2)/(1/2+√5/2)
-(1/(√5/2+5/2))ln[(1/2+√5/2)/(1/2+√5/2)]
+(1/(√5/2-5/2)ln(√5/2-1/2)/(√5/2-1/2)]
=0

收起

∫[0,∞](x^2+1)dx/(x^4-x^2+1)
= -(1/2√5)ln[(1/2+√5/2)/(-1/2+√5/2)] -(1/2√5)ln[(√5/2-1/2)/(1/2+√5/2)-(1/(√5/2+5/2))ln[(1/2+√5/2)/(1/2+√5/2)]+(1/(√5/2-5/2)ln(√5/2-1/2)/(√5/2-1/2)] = 0