The denominator is factorized, so let separate the fraction through partial fraction, hence let

⇒ 5x = A(x – 2)(x + 2) + B(x + 1)(x + 2) + C(x + 1)(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 = – 1 in the above equation, we get

⇒ 5( – 1) = A(( – 1) – 2)(( – 1) + 2) + B(( – 1) + 1)(( – 1) + 2) + C(( – 1) + 1)(( – 1) – 2)

⇒ – 5 = – 3A + 0 + 0

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

⇒ 5( – 2) = A(( – 2) – 2)(( – 2) + 2) + B(( – 2) + 1)(( – 2) + 2) + C(( – 2) + 1)(( – 2) – 2)

⇒ – 10 = 0 + 0 + 4C

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

⇒ 5(2) = A((2) – 2)((2) + 2) + B((2) + 1)((2) + 2) + C((2) + 1)((2) – 2)

⇒ 10 = 0 + 12B + 0

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 + 2 ⇒ 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,