Here the denominator is already factored.

So let

⇒ 1 = A(x – 2)(x – 4) + Bx(x – 4) + Cx(x – 2)……(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 = 0 in the above equation, we get

⇒ 1 = A(0 – 2)(0 – 4) + B(0)(0 – 4) + C(0)(0 – 2)

⇒ 1 = 8A + 0 + 0

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

⇒ 1 = A(2 – 2)(2 – 4) + B(2)(2 – 4) + C(2)(2 – 2)

⇒ 1 = 0 – 4B + 0

Now put x = 4 in equation (ii), we get

⇒ 1 = A(4 – 2)(4 – 4) + B(4)(4 – 4) + C(4)(4 – 2)

⇒ 1 = 0 + 0 + 8C

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 – 4 ⇒ du = dx and z = x – 2 ⇒ dz = dx, so the above equation becomes,

On integrating we get

Substituting back, we get

We will take common, we get

Applying the logarithm rule we can rewrite the above equation as

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

Hence,