I =

As we can see that there is a term of x in numerator and derivative of x² is also 2x. So there is a chance that we can make substitution for –x² + x +2 and I can be reduced to a fundamental integration.

As,

∴ Let, 2x = A(–2x + 1) + B

⇒ 2x = –2Ax + A + B

On comparing both sides –

We have,

–2A = 2 ⇒ A = –1

A + B = 0 ⇒ B = –A = 1

Hence,

I =

∴ I =

Let, I₁ = and I₂ =

Now, I = I₁ + I₂ ….eqn 1

We will solve I₁ and I₂ individually.

As, I₁ =

Let u = 2 + x – x²⇒ du = (–2x + 1)dx

∴ I₁ reduces to

Hence,

I₁ = {∵ }

On substituting value of u, we have:

I₁ = ….eqn 2

As, I₂ = and we don’t have any derivative of function present in denominator. ∴ we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I₂ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in denominator.

∴ I₂ =

⇒ I₂ =

Using: a² + 2ab + b² = (a + b)²

We have:

I₂ =

I₂ matches with

∴ I₂ =

∴ …eqn 3

From eqn 1:

I = I₁ + I₂

Using eqn 2 and eqn 3:

∴ I =