Question

Integrate the following: $ \int {\dfrac{1}{{\left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)}}} $ by using partial fractions.

Answer 1

Hint: Integration by partial fractions: The ratio of two polynomials of a rational function is $ \dfrac{{P\left( x \right)}}{{Q\left( x \right)}} $ , where $ Q\left( x \right) \ne 0 $ , if the degree of the numerator is lower than the degree of the denominator then the fraction is a proper fraction and if the degree of the denominator is lower than the degree of the numerator then the fraction is an improper fraction.

Complete step-by-step answer:
To solve this problem, here, we are using partial fraction decomposition, we will now break the fraction into three different fractions, which will be added together. So, it becomes,
 $ \int {\dfrac{1}{{\left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)}}} dx = \int {\left( {\dfrac{A}{{x - 1}} + \dfrac{B}{{x + 2}} + \dfrac{C}{{x - 3}}} \right)} dx....................................\left( 1 \right) $
Now, on further solving, we get,
 $ \int {\dfrac{1}{{\left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)}}} dx = \int {\left( {\dfrac{{A\left( {x + 2} \right)\left( {x - 3} \right) + B\left( {x - 1} \right)\left( {x - 3} \right) + C\left( {x - 1} \right)\left( {x + 2} \right)}}{{\left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)}}} \right)dx} $
On equating both sides, we get,
 $ \Rightarrow A\left( {x + 2} \right)\left( {x - 3} \right) + B\left( {x - 1} \right)\left( {x - 3} \right) + C\left( {x - 1} \right)\left( {x + 2} \right) = 1.............................................\left( 2 \right) $
Now, we have to find the values $ A,B $ and C and to find these we will substitute any value for x and we have to substitute that number which on substituting makes the factor of a number zero. Let us substitute $ x = 1 $ in equation (2),
 $ A\left( {1 + 2} \right)\left( {1 - 3} \right) + B\left( {1 - 1} \right)\left( {1 - 3} \right) + C\left( {1 - 1} \right)\left( {1 + 2} \right) = 1 $
On further solving, we get,
 $
   \Rightarrow A\left( {1 + 2} \right)\left( {1 - 3} \right) = 1 \\
   \Rightarrow A \times 3 \times \left( { - 2} \right) = 1 \\
   \Rightarrow A = - \dfrac{1}{6} \;
  $
Let us substitute $ x = - 2 $ in equation (2),
 $ A\left( { - 2 + 2} \right)\left( { - 2 - 3} \right) + B\left( { - 2 - 1} \right)\left( { - 2 - 3} \right) + C\left( { - 2 - 1} \right)\left( { - 2 + 2} \right) = 1 $
On further solving, we get,
 $
   \Rightarrow B\left( { - 2 - 1} \right)\left( { - 2 - 3} \right) = 1 \\
   \Rightarrow B \times \left( { - 3} \right) \times \left( { - 5} \right) = 1 \\
   \Rightarrow B = \dfrac{1}{{15}} \;
  $
Let us substitute $ x = 3 $ in equation (2),
 $ A\left( {3 + 2} \right)\left( {3 - 3} \right) + B\left( {3 - 1} \right)\left( {3 - 3} \right) + C\left( {3 - 1} \right)\left( {3 + 2} \right) = 1 $
On further solving, we get,
 $
   \Rightarrow C\left( {3 - 1} \right)\left( {3 + 2} \right) = 1 \\
   \Rightarrow C \times 2 \times 5 = 1 \\
   \Rightarrow C = \dfrac{1}{{10}} \;
  $
Now, we have the values of A, B and C, so, we will substitute it in the equation (1),
 $ \int {\dfrac{1}{{\left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)}}} dx = \int {\left( {\dfrac{{ - \dfrac{1}{6}}}{{x - 1}} + \dfrac{{\dfrac{1}{{15}}}}{{x + 2}} + \dfrac{{\dfrac{1}{{10}}}}{{x - 3}}} \right)} dx $
On further solving, we get,
 $ \Rightarrow - \dfrac{1}{6}\ln \left| {x - 1} \right| + \dfrac{1}{{15}}\ln \left| {x + 2} \right| + \dfrac{1}{{10}}\ln \left| {x - 3} \right| + C $
Hence, this is our answer.
So, the correct answer is “ $ \Rightarrow - \dfrac{1}{6}\ln \left| {x - 1} \right| + \dfrac{1}{{15}}\ln \left| {x + 2} \right| + \dfrac{1}{{10}}\ln \left| {x - 3} \right| + C $ ”.

Note: If we have to integrate $ \int {\left[ {\dfrac{{P\left( x \right)}}{{Q\left( x \right)}}} \right]} dx $ and here $ \dfrac{{P\left( x \right)}}{{Q\left( x \right)}} $ is a proper rational function, in such cases, by using partial fraction decomposition, we can write the integrand as the sum of simpler rational functions. In this problem, we can also find the values of x by equating the denominator of each term with $ 0 $ to find the values of A, B and C.