Consider

Expressing the integral

Let x² + x – 1 = x² + x – 6 + 5

Consider

Factorizing the denominator,

By partial fraction decomposition,

⇒ 1 = A(x + 3) + B(x – 2)

⇒ 1 = Ax + 3A + Bx – 2B

⇒ 1 = (A + B) x + (3A – 2B)

⇒ Then A + B = 0 … (1)

And 3A – 2B = 1 … (2)

Solving (1) and (2),

2 × (1) → 2A + 2B = 0

1 × (2) → 3A – 2B = 1

5A = 1

∴ A = 1/5

Substituting A value in (1),

⇒ A + B = 0

⇒ 1/5 + B = 0

∴ B = -1/5

Thus,

Let x – 2 = u → dx = du

And x + 3 = v → dx = dv.

We know that

Then,

We know that

∴