 Hello friends and how are you all today? The question says, integrate the following rational functions. Now here the function which is given to us is 3x-1 upon x-2 the whole square. So we can write the given rational function as a upon x-2 plus b upon x-2 the whole square. On taking the LCM we have 3x-1 upon x-2 the whole square is equal to a into x-2 plus b upon x-2 the whole square. Now on comparing we get ax is equal to 3x so we have the value of a as 3 and then we have to compare the coefficients of the constants that is minus 2a plus b is equal to minus 1. So on substituting the values of a as 3 we have minus 2 into 3 plus b as minus 1 minus 6 plus b is equal to minus 1 b is equal to plus 5. So we can write the function as 3x-1 upon x-2 the whole square is equal to 3 upon x-2 plus 5 upon x-2 the whole square. On finding out the of integrating both the sides we have 3x-1 upon x-2 the whole square into dx is equal to 3 integral of dx upon x-2 plus 5 integral of dx upon x-2 the whole square. Now we can write it as 3 log mod of x-2 plus 5x-2 raise to the power minus 1 for minus 1 plus c. So the required answer to the session is 3 log mod of x-2 minus 5 upon x-2 plus c. So this completes the session. Hope you understood it and enjoy.