Denominator is factorised, so let separate the fraction through partial fraction, hence let

⇒ ax² + bx + c = A(x – b)(x – c) + B(x – a)(x – c) + C(x – a)(x – b)……(ii)

We need to solve for A, B and C. One way to do this is to pick values for x which will cancel each variable.

Put x = a in the above equation, we get

⇒ a(a)² + b(a) + c = A(a – b)(a – c) + B(a – a)(a – c) + C(a – a)(a – b)

⇒ a³ + ab + c = (a – b)(a – c)A + 0 + 0

Now put x = b in equation (ii), we get

⇒ a(b)² + b(b) + c = A(b – b)(b – c) + B(b – a)(b – c) + C(b – a)(b – b)

⇒ ab² + b² + c = 0 + (b – a)(b – c)B + 0

Now put x = c in equation (ii), we get

We put the values of A, B, and C values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

u = x – a ⇒ du = dx,

y = x – b ⇒ dy = dx and

z = x – c ⇒ dz = dx, so the above equation becomes,

On integrating we get

Substituting back, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,