求(1+4u)/(2u+2-4u^2)的积分,

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求(1+4u)/(2u+2-4u^2)的积分,
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求(1+4u)/(2u+2-4u^2)的积分,
求(1+4u)/(2u+2-4u^2)的积分,

求(1+4u)/(2u+2-4u^2)的积分,
(1+4u)/(2u+2-4u²)
=-(1/2)(1+4u)/[(u-1)(2u+1)]
=A/(u-1)+B/(2u+1)
将上式合并后比较系数可得:A=-5/6,B=-1/3
因此
∫ (1+4u)/(2u+2-4u²) du
=-(5/6)∫1/(u-1)du - (1/3)∫1/(2u+1)du
=-(5/6)ln|u-1| - (1/6)ln|2u+1| + C
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=-[1/(2u+1)+1/(2u-2)+1/(2u^2-u-1)]
故原始积分为:
-{[log(2u+1)]/2+[log(2u-2)]/4+1/6*ln|(u-1/4)/(u+1/4)|}