Evaluate the following integrals –

Let

Let us assume

We know and derivative of a constant is 0.

⇒ x – 2 = λ(2 × 2x^2-1 – 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,