Let

Let us assume

We know and derivative of a constant is 0.

⇒ 4x + 1 = λ(2x^2-1 – 1 – 0) + μ

⇒ 4x + 1 = λ(2x – 1) + μ

⇒ 4x + 1 = 2λx + μ – λ

Comparing the coefficient of x on both sides, we get

2λ = 4 ⇒

Comparing the constant on both sides, we get

μ – λ = 1

⇒ μ – 2 = 1

∴ μ = 3

Hence, we have 4x + 1 = 2(2x – 1) + 3

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

Let

Now, put x² – x – 2 = t

⇒ (2x – 1)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,