Denominator is factorized, so let separate the fraction through partial fraction, hence let

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

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

⇒ 1 = 6A + 0 + 0

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

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

⇒ 1 = 0 – 2B + 0

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

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

⇒ 1 = 0 + 0 + 3C

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

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,