Evaluate the given integral.
 $ \int {\dfrac{{{x^2} + 1}}{{\left( {2x - 1} \right)\left( {{x^2} - 1} \right)}}dx} $

Hint: In this particular question use the concept of partial fraction so after applying partial fraction the function converts into a simple integral so use basic integration formulas to reach the solution of the question.

Complete step-by-step answer:
Given integral:
 $ \int {\dfrac{{{x^2} + 1}}{{\left( {2x - 1} \right)\left( {{x^2} - 1} \right)}}dx} $
Let, $ I = \int {\dfrac{{{x^2} + 1}}{{\left( {2x - 1} \right)\left( {{x^2} - 1} \right)}}dx} $
Now as we know that $ \left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right) $ so use this property in the above integral we have,
 $ \Rightarrow I = \int {\dfrac{{{x^2} + 1}}{{\left( {2x - 1} \right)\left( {x - 1} \right)\left( {x + 1} \right)}}dx} $
Now apply partial fraction we have,
Let, $ \dfrac{{{x^2} + 1}}{{\left( {2x - 1} \right)\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{A}{{\left( {2x - 1} \right)}} + \dfrac{B}{{\left( {x - 1} \right)}} + \dfrac{C}{{\left( {x + 1} \right)}} $
Now simplify we have,
 $ \Rightarrow {x^2} + 1 = A\left( {{x^2} - 1} \right) + B\left( {2x - 1} \right)\left( {x + 1} \right) + C\left( {2x - 1} \right)\left( {x - 1} \right) $
 $ \Rightarrow {x^2} + 1 = A\left( {{x^2} - 1} \right) + B\left( {2{x^2} + x - 1} \right) + C\left( {2{x^2} - 3x + 1} \right) $
Now compare the coefficients of L.H.S and R.H.S we have,
 $ \Rightarrow A + 2B + 2C = 1 $ ................ (1)
 $ \Rightarrow B - 3C = 0 $ ................ (2)
 $ \Rightarrow - A - B + C = 1 $ ............ (3)
Now add equation (2) and (3) we have,
 $ \Rightarrow - A - 2C = 1 $ .............. (4)
Now add (1) and (4) we have,
 $ \Rightarrow 2B = 2 $
 $ \Rightarrow B = 1 $
Now from equation (2) we have,
 $ \Rightarrow 1 - 3C = 0 $
 $ \Rightarrow C = \dfrac{1}{3} $
Now from equation (4) we have,
 $ \Rightarrow - A - 2\left( {\dfrac{1}{3}} \right) = 1 $
 $ \Rightarrow - 1 - 2\left( {\dfrac{1}{3}} \right) = A $
 $ \Rightarrow - 1 - \dfrac{2}{3} = A $
 $ \Rightarrow A = \dfrac{{ - 5}}{3} $
So the integral becomes,
 $ \Rightarrow I = \int {\dfrac{{{x^2} + 1}}{{\left( {2x - 1} \right)\left( {x - 1} \right)\left( {x + 1} \right)}}dx} = \int {\left( {\dfrac{A}{{\left( {2x - 1} \right)}} + \dfrac{B}{{\left( {x - 1} \right)}} + \dfrac{C}{{\left( {x + 1} \right)}}} \right)dx} $
Now substitute the values of A, B and C we have,
 $ \Rightarrow I = \int {\left( {\dfrac{{\dfrac{{ - 5}}{3}}}{{\left( {2x - 1} \right)}} + \dfrac{1}{{\left( {x - 1} \right)}} + \dfrac{{\dfrac{1}{3}}}{{\left( {x + 1} \right)}}} \right)dx} $
Now as we know that ( $ \int {\dfrac{1}{{ax \pm b}}dx} = \dfrac{{\log \left( {ax \pm b} \right)}}{a} + K $ ), where K is some arbitrary integration constant, so use this property in the above integral we have,
 $ \Rightarrow I = \dfrac{{ - 5}}{3}\dfrac{{\log \left( {2x - 1} \right)}}{2} + \dfrac{{\log \left( {x - 1} \right)}}{1} + \dfrac{1}{3}\dfrac{{\log \left( {x + 1} \right)}}{1} + K $
 $ \Rightarrow I = \dfrac{{ - 5}}{6}\log \left( {2x - 1} \right) + \log \left( {x - 1} \right) + \dfrac{1}{3}\log \left( {x + 1} \right) + K $
So this is the required solution of the given integral.

Note: Whenever we face such types of questions the key concept we have to remember is that partial fraction is the key these types of problems only solved by using partial fraction so first apply partial fraction as above then use basic integration formula ( $ \int {\dfrac{1}{{ax \pm b}}dx} = \dfrac{{\log \left( {ax \pm b} \right)}}{a} + K $ ), where K is some arbitrary integration constant so apply this as above we will get the required answer.