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

⇒ 2x – 3 = A(x + 1)(2x + 3) + B(x – 1)(2x + 3) + C(x – 1)(x + 1)……(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

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

⇒ – 5 = 0 – 2B + 0

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

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

⇒ – 1 = 10A + 0 + 0

Now put in equation (ii), we get

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 – 1 ⇒ dy = dx and

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,