Here the denominator is already factored.

So let

⇒ 2x + 1 = A(x – 2) + B(x + 1)……(ii)

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

Put x = 2 in the above equation, we get

⇒ 2(2) + 1 = A(2 – 2) + B(2 + 1)

⇒ 3B = 5

Now put x = – 1 in equation (ii), we get

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

⇒ – 3A = – 1

We put the values of A and B 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 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,