Evaluate the following integral:

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

By equating similar terms, we get

A + C = 0 ⇒ A = – C ………..(iii)

B + D = 0 ⇒ B = – D …………(iv)

3A + C = 2

⇒ 3( – C) + C = 2 (from equation(iii))

⇒ C = – 1

So equation(iii) becomes A = 1

And also 3B + D = 0 (from equation (ii))

⇒ 3( – D) + D = 0 (from equation (iv))

⇒ D = 0

So equation (iv) becomes, B = 0

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

Split up the integral,

Let substitute

so the above equation becomes,

On integrating we get

Substituting back, we get

Applying the logarithm rule we get

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

Hence,