Evaluate the following integrals –

Let

Let us assume

We know and derivative of a constant is 0.

⇒ x + 1 = λ(2 × 2x^2-1 + 0) + μ

⇒ x + 1 = λ(4x) + μ

⇒ x + 1 = 4λx + μ

Comparing the coefficient of x on both sides, we get

4λ = 1 ⇒

Comparing the constant on both sides, we get

μ = 1

Hence, we have

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

Let

Now, put 2x² + 3 = t

⇒ (4x)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,