展开全部
第1个是分部积分
∫ ln(1+t)d[1/(t^2-1)]
= ln(1+t)/(t^2-1) - ∫ [1/(t^2-1)] dln(1+t)
= ln(1+t)/(t^2-1) - ∫ [1/(t^2-1)] [1/(1+t)] dt
第2个是 : partial fraction
1/[(t^2-1)(1+t)]
=1/[(t-1)(t+1)^2]
≡ A/(t-1) +B/(t+1)+C/(t+1)^2
=>
1≡ A(t+1)^2 +B(t-1)(t+1)+C(t-1)
x=1 => A= 1/4
x=-1 => C= -1/2
coef. of x^2
A+B=0
B=-1/4
1/[(t^2-1)(1+t)] ≡ (1/4)[1/(t-1)] -(1/4)[1/(t+1)]-(1/2)[1/(t+1)^2]
∫ ln(1+t)d[1/(t^2-1)]
= ln(1+t)/(t^2-1) - ∫ [1/(t^2-1)] dln(1+t)
= ln(1+t)/(t^2-1) - ∫ [1/(t^2-1)] [1/(1+t)] dt
第2个是 : partial fraction
1/[(t^2-1)(1+t)]
=1/[(t-1)(t+1)^2]
≡ A/(t-1) +B/(t+1)+C/(t+1)^2
=>
1≡ A(t+1)^2 +B(t-1)(t+1)+C(t-1)
x=1 => A= 1/4
x=-1 => C= -1/2
coef. of x^2
A+B=0
B=-1/4
1/[(t^2-1)(1+t)] ≡ (1/4)[1/(t-1)] -(1/4)[1/(t+1)]-(1/2)[1/(t+1)^2]
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