Let

Let us assume

We know and derivative of a constant is 0.

⇒ 2x + 5 = λ(0 – 4 – 3×2x^2-1) + μ

⇒ 2x + 5 = λ(–4 – 6x) + μ

⇒ 2x + 5 = –6λx + μ – 4λ

Comparing the coefficient of x on both sides, we get

–6λ = 2

Comparing the constant on both sides, we get

μ – 4λ = 5

Hence, we have

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

Let

Now, put 10 – 4x – 3x² = t

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

Substituting this value in I₁, we can write

Recall

Let

We can write 10 – 4x – 3x² = –(3x² + 4x – 10)

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,