 Hello and how are you all today? The question says integrate the following rational function. Now here the function which is given to us is 1 minus x square upon x into 1 minus 2x. Now here the integrand is not a proper rational function. So let us first simplify it. So we can write 1 minus x square upon x into 1 minus 2x as 1 minus x square upon x minus 2x square. Now multiplying and dividing the numerator and the denominator by 2 we have in the numerator as 2 minus 2x square upon x minus 2x square in the denominator getting multiplied by 1 by 2. Taking minus sign common in the numerator and denominator we have minus 1 upon minus 2 into 2x square minus 2 upon 2x square minus x. These two minus signs will get cancelled. Now further we can write now on adding and subtracting x in the numerator we have 2x square minus x minus sorry plus x minus 2 whole divided by 2x square minus x. That can be further written as 1 upon 2 into 1 plus x minus 2 upon 2x square minus x. Now let us simplify the fraction. Now we need to simplify this fraction separately. So we have x minus 2 upon 2x square minus x. Now it can be written as x minus 2 upon taking sign x common we are left with 2x minus 1 in the denominator and this can be written as a upon x plus b upon 2x minus 1. Further we have a into 2x minus 1 plus bx upon x into 2x minus 1 and this is equal to x minus 2 upon 2x square minus x. So on comparing we have the coefficients of x that is 2a plus b equal to 1 and the constants that is minus a equal to minus 2. So we have the value of a as 2 and therefore the value of b substituted here as 2 we have the value of b as minus 3. So we can write this function as 1 upon 2 plus 1 upon 2 into and in the bracket now we have a upon x and a is 2 plus b that is minus 3 upon 2x minus 1. Right. So further let us simplify it from here on. So we have now the fraction that we had in the starting of the session that was 1 minus x square upon x into 1 minus 2x is now equal to 1 upon 2 plus 1 upon x minus 3 upon 2 2x minus 1. Now integrating both the sides we have integral of 1 minus x square upon x into 1 minus 2x dx is equal to integral of 1 by 2 dx plus integral of dx upon x minus minus 3 by 2 dx upon 2x minus 1. So we have the answer as x upon 2 plus log mod x minus 3 upon 2 log 2x minus 1 divided by the derivative of 2x minus 1 that is 2 plus c. Right. So we can write the answer as x upon 2 plus log of mod x minus 3 upon 4 log log mod of Now we know that the value of a mod is always positive. So we can rearrange the terms inside also. So we have minus 3 by 4 log 1 minus 2x plus c and this is the required answer to this question. Hope you enjoy it and have a good day.