积分∫dx/e^x-e^-x=?
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简单计算一下即可，答案如图所示
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∫ dx/(e^x - e^-x)
= ∫ e^x/(e^(2x) - 1) dx
= ∫ d(e^x)/(e^(2x)-1)
= ∫ du/(u²-1),u=e^x,下一步因式分解,see the Notes below.
= -(1/2)∫ du/(1+u) + (1/2)∫ du/(u-1)
= -(1/2)ln|1+u| + (1/2)ln|u-1| + C
= (1/2)ln|(u-1)/(u+1)| + C
= (1/2)ln|(e^x-1)/(e^x+1)| + C
Notes：
1/(u²-1) = 1/((u+1)(u-1)) = A/(u+1) + B/(u-1)
1 = A(u-1) + B(u+1)
1 = (A+B)u + (B-A)
A+B=0 => B=-A
B-A=1 => -A-A=1 => A=-1/2
B=1/2
= ∫ e^x/(e^(2x) - 1) dx
= ∫ d(e^x)/(e^(2x)-1)
= ∫ du/(u²-1),u=e^x,下一步因式分解,see the Notes below.
= -(1/2)∫ du/(1+u) + (1/2)∫ du/(u-1)
= -(1/2)ln|1+u| + (1/2)ln|u-1| + C
= (1/2)ln|(u-1)/(u+1)| + C
= (1/2)ln|(e^x-1)/(e^x+1)| + C
Notes：
1/(u²-1) = 1/((u+1)(u-1)) = A/(u+1) + B/(u-1)
1 = A(u-1) + B(u+1)
1 = (A+B)u + (B-A)
A+B=0 => B=-A
B-A=1 => -A-A=1 => A=-1/2
B=1/2
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