∫(2x+6)/[(x^2-1)(x^2+1)^2] dx

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∫(2x+6)/[(x^2-1)(x^2+1)^2] dx

∫(2x+6)/[(x^2-1)(x^2+1)^2] dx
∫(2x+6)/[(x^2-1)(x^2+1)^2] dx

∫(2x+6)/[(x^2-1)(x^2+1)^2] dx
这题要利用有理函数分解原理,将原式拆分
原式=(1/2)∫(x+1)/(x^2-1)dx-(1/2)∫(x+9)/(x^2+1)dx-∫(x+1)/(x^2+1)^2dx
=(1/2)∫1/(x-1)d(x-1)-(1/2)∫1/(x^2+1)d(x^2+1)-(9/2)∫1/(x^2+1)dx-∫1/(x^2+1)^2d(x^2+1)-∫1/(x^2+1)^2dx
【除了最后一个积分其他都是常见常用积分了,可自己计算,下面来计算最后一个积分】
对于∫1/(x^2+1)^2dx
令x=tant,t∈(-π/2,π/2)
该积分=∫(cost)^2dt
=∫(1+cos2t)/2dt
=(1/2)∫dt+(1/4)∫cos2td(2t)
=(1/2)t+(1/4)sin2t+C
=(1/2)t+(1/2)sintcost+C
根据tant=x/1,作辅助三角形,可得
sint=x/√(x^2+1)
cost=1/√(x^2+1)
将它们回代即可.

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