Evaluate the following integral:

First we simplify numerator, we get






Now we will factorize denominator by splitting the middle term, we get






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




5 = A(x – 2) + B(x + 3)……(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 x = 2 in the above equation, we get


5 = A(2 – 2) + B(2 + 3)


5 = 0 + 5B


B = 1


Now put x = – 3 in equation (ii), we get


5 = A(( – 3) – 2) + B(( – 3) + 3)


5 = – 5A


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 + 3 du = dx and z = x – 2 dz = dx, so the above equation becomes,



On integrating we get


x – log|u| + log|z| + C


Substituting back, we get


x – log|x + 3| + log|x – 2| + C


Applying the logarithm rule, we can rewrite the above equation as



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


Hence,



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