Evaluate the following integral:

Question

Philoid · Accepted Answer

First we simplify numerator, we get

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

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 x = 1 in the above equation, we get

⇒ 2 = A(1 – 1) + B(1 + 1)

⇒ 2 = 0 + 2B

⇒ B = 1

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

⇒ 2 = A(( – 1) – 1) + B(( – 1) + 1)

⇒ 2 = – 2A + 0

⇒ A = – 1

We put the values of A and B 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 and z = x – 1 ⇒ dz = dx, 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,