 Hello and welcome to the session. Let us discuss the following question. Question says, choose the correct answer integral x dx upon x minus 1 multiplied by x minus 2 equals a log x minus 1 whole square upon x minus 2 plus c b log x minus 2 whole square upon x minus 1 plus c c log of x minus 1 upon x minus 2 whole square plus c b log of x minus 1 multiplied by x minus 2 plus c. Let us now start with the solution. Now we have to find the integral of x dx upon x minus 1 multiplied by x minus 2. Clearly we can see integrant is a proper rational function. So we can write x upon x minus 1 multiplied by x minus 2 is equal to a upon x minus 1 plus b upon x minus 2. Now this can be further written as x is equal to a multiplied by x minus 2 plus b multiplied by x minus 1. Now we get x is equal to a x minus 2 a plus b x minus b. Now equating the coefficients of x and constant term we get a plus b is equal to 1 and minus 2 a minus b is equal to 0. Now solving these two equations we get a is equal to minus 1 and b is equal to 2. Now substituting values of a and b in this expression we get x upon x minus 1 multiplied by x minus 2 is equal to minus 1 upon x minus 1 plus 2 upon x minus 2. Therefore integral x dx upon x minus 1 multiplied by x minus 2 is equal to integral minus dx upon x minus 1 plus 2 multiplied by integral dx upon x minus 2. Now this expression is further equal to minus 1 or we can say minus integral dx upon x minus 1 plus 2 multiplied by integral dx upon x minus 2. Now let us assume that this integral is i1 and this integral is i2 and we can write it as minus i1 plus 2 i2. First upon let us consider i1 you know i1 is equal to dx upon x minus 1. Now put x minus 1 is equal to t. Now differentiating both the sides with respect to x we get dx is equal to dt. Now we get i1 is equal to integral of dt upon t. Now this is further equal to log t plus c. To find this integral we have used this formula of integral here x has been replaced by t. Now after finding the integral we will substitute value of t here we get log of x minus 1 plus c. Similarly we can find i2 we know i2 is equal to integral of dx upon x minus 2. Using this same method we can find integral of dx upon x minus 2. So we can write that integral of dx upon x minus 2 is equal to log of x minus 2 plus c. Now we will substitute corresponding values of i1 and i2 in this expression. Now we get integral of x dx upon x minus 1 multiplied by x minus 2 is equal to minus log of x minus 1 plus 2 multiplied by log of x minus 2 plus c. Now this expression can be further written as log of x minus 2 whole square upon x minus 1 plus c. Using these laws we get this expression is equal to this expression. So our correct answer is b. This is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.