Evaluate the following integral:

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

⇒ 5x² – 1 = A(x – 1)(x + 1) + Bx(x + 1) + Cx(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 = 0 in the above equation, we get

⇒ 5(0)² – 1 = A(0 – 1)(0 + 1) + B(0)(0 + 1) + C(0)(0 – 1)

⇒ A = 1

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

⇒ 5(1)² – 1 = A(1 – 1)(1 + 1) + B(1)(1 + 1) + C(1)(1 – 1)

⇒ 4 = 0 + 2B + 0

⇒ B = 2

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

⇒ 5( – 1)² – 1 = A( – 1 – 1)( – 1 + 1) + B( – 1)( – 1 + 1) + C( – 1)( – 1 – 1)

⇒ 4 = 0 + 0 + 2C

⇒ C = 2

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 logarithm rule, we get

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

Hence,