 Welcome to the session, I am Deepika here. Let's discuss the question which says, integrate the following rational function, 2x minus 3 upon x square minus 1 into 2x plus c. Now we know that if our rational function is plus qx plus x minus c, then fraction is x minus a plus b plus c determined suitably. So this is a key idea behind a key idea to solve the above question. So let's start the solution. Integrate the rational function 2x minus 3 upon the function. Therefore, according to our key idea or by using the form of partial fraction, we can write x minus 1 plus x minus 1 which is x square minus 1. Therefore, 2x minus 3 minus d. The coefficient of x cube equation as number 1, the coefficient of equation as number 2, the coefficient of let us give this as number 3, let us give this as equation 2 and equation 4 we get now 2 is equation 4 is minus as number 5. The value of equation 5 is 1 is equal to this 1 upon the values of equation 3 to 1 by 10 minus 0 is equal to this implies this 1 by 2 which is equal to 5 by 2. To obtain the values of equation 2 was 2 equal to minus 1 by 10 1 by 5 is equal to this 24 upon 5 2 minus 1 upon 10, c is equal to 0 and d is equal to minus 24 upon 5. This is minus 1 over 10 into x minus 1 plus c is equal to 0 and d is minus 24 upon 5. So, this is minus 24 upon minus 1 into 2 integral of 1 over x plus 1 dx minus 1 by 10 into integral of 1 over x minus 1 dx 4 upon 5 into integral of 1 over 2x plus 3 dx. The process of differentiation and integration are inverses of each other derivative of log of x plus 1 is equal to 1 over x log mod of x plus 1 minus 1 by 10 log of mod x minus 1 into log mod of 2x plus 3 plus c the question is mod x plus 1 minus 1 by 10 log mod x minus 1 minus 12 upon 5 x plus 3 plus c. I hope the solution is clear to you. Bye and take care.