Evaluate the following integrals –

Let


Let us assume




We know and derivative of a constant is 0.


2x – 5 = λ(2x2-1 – 4 + 0) + μ


2x – 5 = λ(2x – 4) + μ


2x – 5 = 2λx + μ – 4λ


Comparing the coefficient of x on both sides, we get


2λ = 2 λ = 1


Comparing the constant on both sides, we get


μ – 4λ = –5


μ – 4(1) = –5


μ – 4 = –5


μ = –1


Hence, we have


Substituting this value in I, we can write the integral as





Let


Now, put x2 – 4x + 3 = t


(2x – 4)dx = dt (Differentiating both sides)


Substituting this value in I1, we can write




Recall






Let


We can write x2 – 4x + 3 = x2 – 2(x)(2) + 22 – 22 + 3


x2 – 4x + 3 = (x – 2)2 – 4 + 3


x2 – 4x + 3 = (x – 2)2 – 1


x2 – 4x + 3 = (x – 2)2 – 12


Hence, we can write I2 as



Recall





Substituting I1 and I2 in I, we get



Thus,


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