Let

Let us assume

We know and derivative of a constant is 0.

⇒ 3x + 1 = λ(0 – 3 – 2×2x^2-1) + μ

⇒ 3x + 1 = λ(–3 – 4x) + μ

⇒ 3x + 1 = –4λx + μ – 3λ

Comparing the coefficient of x on both sides, we get

–4λ = 3

Comparing the constant on both sides, we get

μ – 3λ = 1

Hence, we have

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

Let

Now, put 4 – 3x – 2x² = t

⇒ (–3 – 4x)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

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

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,