Let

Let us assume

We know and derivative of a constant is 0.

⇒ 2x – 5 = λ(0 + 3 – 2x^2-1) + μ

⇒ 2x – 5 = λ(3 – 2x) + μ

⇒ 2x – 5 = –2λx + 3λ + μ

Comparing the coefficient of x on both sides, we get

–2λ = 2 ⇒ λ = –1

Comparing the constant on both sides, we get

3λ + μ = –5

⇒ 3(–1) + μ = –5

⇒ –3 + μ = –5

∴ μ = –2

Hence, we have

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

Let

Now, put 2 + 3x – x² = t

⇒ (3 – 2x)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,