Evaluate the following integrals –

Let

Let us assume

We know and derivative of a constant is 0.

⇒ 2x – 5 = λ(2x^2-1 – 4 + 0) + μ

⇒ 2x – 5 = λ(2x – 4) + μ

⇒ 2x – 5 = 2λx + μ – 4λ

Comparing the coefficient of x on both sides, we get

2λ = 2 ⇒ λ = 1

Comparing the constant on both sides, we get

μ – 4λ = –5

⇒ μ – 4(1) = –5

⇒ μ – 4 = –5

∴ μ = –1

Hence, we have

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

Let

Now, put x² – 4x + 3 = t

⇒ (2x – 4)dx = dt (Differentiating both sides)

Substituting this value in I₁, we can write

Recall

Let

We can write x² – 4x + 3 = x² – 2(x)(2) + 2² – 2² + 3

⇒ x² – 4x + 3 = (x – 2)² – 4 + 3

⇒ x² – 4x + 3 = (x – 2)² – 1

⇒ x² – 4x + 3 = (x – 2)² – 1²

Hence, we can write I₂ as

Recall

Substituting I₁ and I₂ in I, we get

Thus,