∫01 x−1x2−1dx
ℓn(2)
ℓn(6) – ℓn(3)
ℓn(3) – ℓn(6)
– ℓn(2)
∫01 x−1x2−1=∫01 x−1(x+1)(x−1)=∫01 1(x+1)dx=ℓn(x+1)01=ℓn2