 Welcome to the session I am Deepika here. Let's discuss a question which says integrate the following rational function x upon x square plus 1 into x minus 1. Now we know that it is always possible to write the integrand as a sum of simpler rational functions by a method called partial fraction decomposition. So we will integrate this question by partial fractions. Now we know that if the form of the rational function is p x square plus q x plus r upon into x square plus b x plus c where plus b x plus c cannot be factorized further than third of the partial fraction is over x minus a plus upon x square plus b x plus c where two numbers will be determined suitably. So this is a key idea behind our question. We will take the help of this key idea to solve the above question. So let's start the solution. Now we have to integrate the rational function x upon x square plus 1 into x minus 1. The given integrand is a proper rational function according to our key idea or by using the form of partial fraction we have square plus 1 into x minus 1 is equal to x plus b upon x square plus 1 p upon x minus 1 where a, b and l numbers will be determined suitably. So this is equal to ax plus b x minus 1 plus c into x square plus 1. So this implies x is equal to square minus ax plus bx minus b plus cx square plus c the coefficients of x square the constant term is equal to 0. Let us give this equation as number one. Let us give this equation as number two equal to zero. Let us give this equation as number three. Now we will solve these equations. So from equation three we have b is equal to c. Now on substituting b is equal to c in equation two we have minus a plus c is equal to 1. Let us give this equation as number four. Now from one we have a plus equal to zero. Now on adding we get c is equal to 1. So this implies a is equal to 1 by 2. From one we have a plus c is equal to zero. Now on substituting the value of c we get a plus 1 by 2 is equal to zero. So this implies a is equal to minus 1 by 2. We have a is equal to minus 1 by 2 equal to c is equal to 1 by 2. Thus the integrand is given by x square plus 1 into x minus 1 is equal to a x plus b. Now a is minus 1 by 2. So this is minus 1 by 2 x plus b which is 1 by 2 upon x square plus 1 dx is equal to 1 by 2 plus 1 by 2 into integral of 1 over x minus 1. Can be written as is equal to integral of integral plus 1 into dx plus 1 by 1 over x square plus 1 into dx plus 1 by 2 into 1 over x minus 1 into dx into integral of is equal to dt. Therefore integral of x upon x square plus 1 dx is equal to minus 1 by 4 minus 1 by 4. This is again equal to minus 1 by 4 log x square plus 1 dx is equal to minus 1 by 4 log of that the process of differentiation and integration are inverses of each other because derivative of tan inverse x to 1 over 1 plus x square plus 1 into x minus 1 into integral of x upon x square plus 1 dx is equal to minus 1 by 4 log of x square plus 1 square plus 1 into x minus 1 dx log of 1 over x square plus 1 dx and which is 1 by 2 tan 1 by 2 into integral of 1 over x minus 1 dx which is equal to 1 by 2 into log of a question is plus 1 by 2 solution is clear to you.