I = ∫_1 ^2 ((2x^2 )/( (√((2x−1)(2x+2))))) dx = 2∫_1 ^2 (x^2 /( (√((2x−1)(2x+2))))) dx =^(t = 2x) (1/4)∫_2 ^( 4) (t^2 /( (√((t−1)(t+2))))) dt Let ξ(t−1) = (√((t−1)(t+2))) ⇒ t = ((2+ξ^2 )/(ξ^2 −1)) ⇒ dt = −((6ξ)/((ξ^2 −1)^2 )) dξ I = (1/4)∫_2 ^( (√2)) (((((2+ξ^2 )/(ξ^2 −1)))^2 )/(ξ(((2+ξ^2 )/(ξ^2 −1))−1)))(−((6ξ)/((ξ^2 −1)^2 ))) dξ I = (1/4)∫_(√2) ^( 2) ((((2+ξ^2 )^2 )/((ξ^2 −1)^2 ))/(3/(ξ^2 −1)))((6/((ξ^2 −1)^2 ))) dξ I = (1/2)∫_(√2) ^( 2) (((2+ξ^2 )^2 )/((ξ^2 −1)^3 )) dξ Q(x) = (ξ^2 −1)^3 ⇒ Q′(x) = 6ξ(ξ^2 −1)^2 ⇒ Q_1 (x) = (ξ^2 −1)^2 & Q_2 (x) = ξ^2 −1 ∫ (((2+ξ^2 )^2 )/((ξ^2 −1)^3 )) dξ = ((Aξ^3 +Bξ^2 +Cξ+D)/((ξ^2 −1)^2 )) + ∫ ((Eξ+F)/(ξ^2 −1)) dξ ((4+4ξ^2 +ξ^4 )/((ξ^2 −1)^3 )) = (((ξ^2 −1)(3Aξ^2 +2Bξ+C)−4ξ(Aξ^3 +Bξ^2 +Cξ+D))/((ξ^2 −1)^3 )) + ((Eξ+F)/(ξ^2 −1)) ⇒ A = (3/8), B = D = E = 0, C = −((21)/8), F = ((11)/8) I = (1/2)[((3ξ^3 −21ξ)/(8(ξ^2 −1)^2 ))∣_(√2) ^2 + ((11)/8)∫_(√2) ^( 2) (1/(ξ^2 −1))∣ dξ] = (1/2)[((3ξ^3 −21ξ)/(8(ξ^2 −1)^2 ))∣_(√2) ^2 + ((11)/8)((1/2)ln(∣((ξ−1)/(ξ+1))∣))∣_(√2) ^2 ] = (1/2)[−(1/4) + ((15(√2))/8) + ((11)/8)(((−ln(3)+ln(3−(√2)))/2))] = (1/2)[−(1/4) + ((15(√2))/8) + ((11)/(16))ln(((3−(√2))/3))] = ((11)/(32))ln(((3−(√2))/3)) +((15(√2))/(16)) −(1/8) determinant (((∫_1 ^2 ((2x^2 )/( (√((2x−1)(2x+2))))) dx = ((11)/(32))ln(((3−(√2))/3)) + ((15(√2))/(16)) − (1/8)))) Forgot it′s Question No. I kind of missed the past times...