Denominator is already factorized, so let

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 = 1 in the above equation, we get

= A(1 – 2)(1 – 3) + B(1 – 1)(1 – 3) + C(1 – 1)(1 – 2)

⇒ 1 = 2A + 0 + 0

Now put x = 2 in equation (ii), we get

= A(2 – 2)(2 – 3) + B(2 – 1)(2 – 3) + C(2 – 1)(2 – 2)

⇒ 4 = 0 – B + 0

⇒ B = – 4

Now put x = 3 in equation (ii), we get

= A(3 – 2)(3 – 3) + B(3 – 1)(3 – 3) + C(3 – 1)(3 – 2)

⇒ 9 = 0 + 0 + 2C

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 – 1 ⇒ du = dx, y = x – 2 ⇒ dy = dx and z = x – 3 ⇒ 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,