 Hello friends and how are you all today? Let us integrate the following rational function. Now here in this question the integrant is a proper rational function. So we will be solving it by the method of partial fractions. So let us start with our solution. Here we can write 3x-1 upon x-1 into x-2 into x-3 equals to a upon x-1 plus b upon x-2 plus c upon x-3. Now taking LCM we have 3x-1 upon x-1 into x-2 into x-3 written as e x-2 into x-3 plus b x-1 into x-3 plus c x-1 into x-2 the whole upon x-1 into x-2 into x-3. Now on solving the brackets we have it equal to a into x-2 minus 5x plus 6 plus b into these 2 can be solved and written as x-2 minus 4x plus 3 plus c into x-2 minus 3x plus 2 the whole divided by x-1 to x-2 into x-3 and these are all equal to 3x-1 upon x-1 into x-2 into x-3. Now we need to compare the various terms like the coefficient of powers of x and constants. So let us rearrange the right hand side and write x-2 into a plus b plus c plus x into minus 5a minus 4b minus 3c and plus 6a plus 3b plus 2c and the numerator is equal to 3x minus 1 right. Now comparing the coefficient of x-2 on both the sides we get since in the left hand side there is no term in x-2 so we have a plus b plus c equals to 0 and then comparing the coefficients of x we have minus 5a minus 4b minus 3c equals to 3 let this be the first equation this be the second and then comparing the coefficients we have 6a plus 3b plus 2c equal to 1 let this be the third equation. Now multiplying the first equation by 3 we have 3a plus 3b plus 3c equals to 0 let this be the fourth equation. Now on adding the fourth equation and the second equation we get 3a plus 3b plus 3c equal to 0 and the second equation is minus 5a minus 4b minus 3c equal to 3. Now we need to add all these terms and on adding we have minus 2a minus b equal to 3 let this be the fifth equation. Now further on multiplying the first equation by 2 and subtracting the resultant equation from the sixth equation sorry from the third equation we get. Now the third equation was known to us as 6a plus 3b plus 2c equal to minus 1 now subtracting the first equation after multiplying it by 2 we have 2a plus 2b plus 2c equal to 0. So we have the result as 4a minus plus b equal to minus 1 that means let's name this equation as the sixth equation and therefore we can have the value of b as minus 1 minus minus 4a. This is b the seventh equation now substituting the value of b from the seventh equation in the fifth equation that is minus 2a minus b is equal to 3 we get minus 2a minus b that is minus 1 minus 4a equal to 0. So we have minus 2a plus 1 plus 4a equal to 3 that gives us 2a is equal to 2 that further implies the value of a is equal to 1. Now substituting the value of a in the seventh equation we have minus 1 minus 4 the value of a is 1 is equal to 3 that is minus 1 minus 4 it was b not 3. So minus 1 minus 4 is equal to b that further implies the value of b is equal to minus 5. Now we need to find out the value of c so we have a plus b plus c equal to 0 that was our first equation let us substitute the values. So we have 1 plus minus 5 plus c equal to 0 that further implies 1 minus 5 plus c is equal to 0 that is minus 4 plus c is equal to 0 that implies the value of c is equal to 4. So now we have the value of a b and c so we can write 3x minus 1 upon x minus 1 into x minus 2 into x minus 3 equal to a that was 1 upon x minus 1 b that was minus 5 upon x minus 2 plus c upon that is 4 x minus 3. Now integrating both the sides 3x minus 1 upon x minus 1 into x minus 2 x minus 3 into dx is equal to integration of 1 upon x minus 1 minus 5 upon x minus 2 plus 4 upon x minus 3 to dx. Now so now it can be written as equal to integral of 1 upon x minus 1 into dx minus 5 integral of 1 over x minus 2 dx plus 4 integral of 1 upon x minus 3 dx. Now we know that integral of 1 upon x dx is equal to log mod of x plus c right so we can write it equal to log mod of x minus 1 minus 5 log x minus 2 plus 4 log x minus 3 plus c. So the required answer to the session is log mod x minus 1 minus 5 log x minus 2 plus log plus 4 log x minus 3 plus c right so this is the required answer hope you enjoyed and understood it well have a very nice day ahead.