Evaluate the following integral:
Let substitute, so the given equation becomes
Factorizing the denominator, we get
The denominator is factorized, so let separate the fraction through partial fraction, hence let
⇒ 1 = A(3u + 2) + B(2u + 1)……(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 in the above equation, we get
Now put in equation (ii), we get
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 = 2u + 1 ⇒ dz = 2du and y = 3u + 2⇒ dy = 3du 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,