 Hello and welcome to the session. Let us discuss the following question. Question says, integrate the following functions. Given function is 6x plus 7 upon square root of x minus 5 multiplied by x minus 4. Let us now start with the solution. Now we have to find integral of the function 6x plus 7 upon square root of x minus 5 multiplied by x minus 4 with respect to x. First of all, let us find out the product x minus 5 multiplied by x minus 4. It is equal to x square minus 9x plus 20. Now let us express 6x plus 7 as a multiplied by d by dx of x square minus 9x plus 20 plus b. Now we get 6x plus 7 is equal to a multiplied by 2x minus 9. We know differentiating this expression with respect to x. We get 2x minus 9. And we will write plus b as it is. Now equating the coefficients of x on both the sides we get a is equal to 6 upon 2 that is 3. Now equating the constant terms on both the sides we get minus 9a plus b is equal to 7. Now substituting value of a is equal to 3 in this expression we get minus 9 multiplied by 3 plus b is equal to 7. Now this further implies b is equal to 7 plus 27. Now we get b is equal to 34. Now substituting values of a and b in this expression we get. Now we have to find out this integral. Now this integral can be written as integral of 6x plus 7 upon square root of x square minus 9x plus 20. Multiplying these two brackets we get these three terms and here we will write dx. Now substituting this value for 6x plus 7 in this integral we get 3 multiplied by integral of 2x minus 9 upon square root of x square minus 9x plus 20 dx plus 34 multiplied by integral dx upon square root of x square minus 9x plus 20. Now let us name these integrals as i1 and i2. So we can write this expression as 3i1 plus 34 i2 First of all let us find out i1. You know i1 is equal to integral of 2x minus 9 upon square root of x square minus 9x plus 20 dx. Now put x square minus 9x plus 20 is equal to t. Now differentiating both the sides with respect to x we get 2x minus 9 is equal to dt upon dx. Now this further implies 2x minus 9 dx is equal to dt. Now we can write i1 as integral of dt upon root t. We know this expression is equal to t and 2x minus 9 dx is equal to dt. Now this integral is further equal to 2 root t plus c. Now 1 upon root t means t raised to the power minus 1 upon 2. To find the integral of t raised to the power minus 1 upon 2 we will use this formula of integration. Now substituting this expression for t here we get 2 multiplied by square root of x square minus 9x plus 20 plus c. Now let us find out i2. You know i2 is equal to integral of dx upon square root of x square minus 9x plus 20. Now x square minus 9x plus 20 is equal to whole square of x minus 9 upon 2 minus 1 upon 2 square. Now we can write this integral as i2 is equal to integral of dx upon square root of x minus 9 upon 2 whole square minus square of 1 upon 2. Now put x minus 9 upon 2 equal to t. Differentiating both sides with respect to x we get dx is equal to dt. Now we get i2 is equal to integral of dt upon square root of t square minus square of 1 upon 2. We know dx is equal to dt and x minus 9 upon 2 is equal to t. Now this integral is further equal to log of t plus square root of t square minus square of 1 upon 2 plus c. For finding this integral we have used this formula. Here x has been replaced by t and a has been replaced by 1 upon 2. Now substituting x minus 9 upon 2 for t in this expression we get i2 is equal to log of x minus 9 upon 2 plus square root of x minus 9 upon 2 whole square minus square of 1 upon 2 plus c. Now i2 can be further written as log of x minus 9 upon 2 plus x square minus 9x plus 20 plus c. Now substituting values of i1 and i2 in this expression we get integral of 6x plus 7 dx upon square root of x minus 5 multiplied by x minus 4 is equal to 3 multiplied by 2 square root of x square minus 9x plus 20 plus 34 multiplied by log of x minus 9 upon 2 plus square root of x square minus 9x plus 20 plus c. Now this can be further written as 6 multiplied by square root of x square minus 9x plus 20 plus 34 multiplied by log of x minus 9 upon 2 plus square root of x square minus 9x plus 20 plus c. Remember that c is the constant of integration. So this is our required answer. Take care and have a nice day.