Evaluate the following integrals –

Let


Let us assume




We know and derivative of a constant is 0.


x – 2 = λ(2 × 2x2-1 – 6 – 0) + μ


x – 2 = λ(4x – 6) + μ


x – 2 = 4λx + μ – 6λ


Comparing the coefficient of x on both sides, we get


4λ = 1


Comparing the constant on both sides, we get


μ – 6λ = –2





Hence, we have


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






Let


Now, put 2x2 – 6x + 5 = t


(4x – 6)dx = dt (Differentiating both sides)


Substituting this value in I1, we can write




Recall







Let


We can write






Hence, we can write I2 as





Recall






Substituting I1 and I2 in I, we get



Thus,


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