First we simplify numerator, we get

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

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

⇒ 5(1) + 1 = A(1 – 1) + B(1 + 2)

⇒ 6 = 0 + 3B

⇒ B = 2

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

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

⇒ – 9 = – 3A + 0

⇒ A = 3

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 + 2 ⇒ du = dx and z = x – 1 ⇒ 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,