Let substitute, so the given equation becomes

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

⇒ 1 = A(2 + u) + Bu……(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 u = – 2 in above equation, we get

⇒ 1 = A(2 + ( – 2)) + B( – 2)

⇒ 1 = – 2B

Now put u = 0 in equation (ii), we get

⇒ 1 = A(2 + 0) + B(0)

⇒ 1 = 2A + 0

We put the values of A and B values back into our partial fractions in equation (ii) and replace this as the integrand. We get

Split up the integral,

Let substitute

z = 2 + u ⇒ dz = du, so the above equation becomes,

On integrating we get

Substituting back the value of z, we get

Now substitute back the value of u, we get

Applying the rules of logarithm we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,