Evaluate the following integral:

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

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 = 0 in the above equation, we get

⇒ 0² + 1 = A(0 – 1)(0 + 1) + B(0)(0 + 1) + C(0)(0 – 1)

⇒ 1 = – A + 0 + 0

⇒ A = – 1

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

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

⇒ 2 = 0 + 0 + C

⇒ C = 1

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

⇒ 1² + 1 = A(1 – 1)(1 + 1) + B(1)(1 + 1) + C(1)(1 – 1)

⇒ 2 = 0 + 2B + 0

⇒ B = 1

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, so the above equation becomes,

On integrating we get

Substituting back, 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,