 Hello friends and how are you all today? My name is Priyanka and the question says, integrate the following rational function. Now here the rational function which is given to us is 1 upon x raised to the power 4 minus 1. Now we can factorize the denominator and write it as 1 upon x minus 1 into x plus 1 which is x square minus 1 into x square plus 1. Now the integrand is a proper rational fraction which can be integrated effectively by the principle of rational fractions. So therefore we get 1 upon x raised to the power 4 minus 1 is equal to a upon x minus 1 plus b upon x plus 1 plus since this is quadratic we will have cx plus d in the numerator upon x square plus 1. Now taking the LCM of the right hand side we get and equating the numerator also we will have 1 is equal to a into x plus 1 into x square plus 1 plus b into x minus 1 into x square plus 1 plus cx plus d into x square minus 1 right. So we have 1 is equal to let us simplify it further x cube plus x square plus x plus 1 plus b into x cube minus x square plus x plus 1 plus c into x cube minus x plus d into x square minus 1. Now on equating the coefficients of x cube, x square, x and constants we have now for x cube when we are equating the coefficient of x cube we have a plus b plus c is equal to 0 when we are equating the coefficient of x square we have a minus b plus d is equal to 0 when we are doing it for x then we have a plus b minus c is equal to 0 and then for the constants it is a plus b minus d is equal to 1 right. Now on solving these four equations we have the values of a, b, c and d equal to the value of a is 1 by 4 value of b is minus 1 by 4 value of c is obtained as 0 and value of d is minus 1 by 2 right. Now substituting the values in the above equation we get that is the value of a, b, c and d integral of 1 upon x4 minus 1 we have integrated both the sides also is equal to since the value of c is 0 we will not be taking that term so it is 1 upon 4 into x minus 1 minus 1 upon 4 into x plus 1 minus 1 upon 2 into x square plus 1 into dx. So taking integration sign separately with each term we have integral dx upon 4 into x minus 1 minus integral dx upon 4 into x plus 1 minus integral dx upon 2 x square plus 1. Now taking constant terms outside we get 1 upon 4 integral dx upon x minus 1 minus 1 upon 4 integral dx upon x plus 1 minus 1 upon 2 integral dx upon x square plus 1. Since we know that integral of 1 upon x dx is equal to log x plus c so we have equal to 1 upon 4 log mod x minus 1 minus 1 upon 4 log mod x plus 1 minus 1 upon 2 now integral of dx upon x square plus 1 is tan inverse x and taking all the constants we will denote it by c. So the required answer is 1 upon 4 log these two terms can be combined and written as mod of x minus 1 upon mod of x plus 1 minus 1 upon 2 tan inverse x plus c and this is the required answer for the session. Hope you understood the whole concept well and have a nice day ahead.