# 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).

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