 Hello friends and how are you all today? The question says integrate the following functional functions. So here we are given x square plus x plus 1 upon x square into x plus 2. Now here we will be using integration by partial fractions that says that if we have a function that is px square plus qx plus r upon x minus a the whole square into x minus b and we can write it as a upon x minus a plus b upon x minus a the whole square plus c upon x minus b. The knowledge of this part of the integration is the key idea for the question. Now we can write the given function as x square plus x plus 1 upon x square into x plus 2 equal to a upon x plus b upon x square plus c upon x plus 2. So taking the LCM we have x square plus x plus 1 upon x square into x plus 2 written as a into x into x plus 2 plus b into x plus 2 plus c into x square. We will divide it by x square into x plus 2. Now on further simplifying we have x square plus x plus 1 upon x square into x plus 2 is equal to a x square plus 2x plus b x plus 2 plus c x square whole divided by x square into x plus 2. Now on comparing the coefficients we have one in the right hand side and the coefficients of x square that is a plus c. Again one on the right hand side and the coefficient of x that is 2a plus b and then we have the constants that is equal to 1 and we have now on simplifying these three equations here we can get the value of b as 1 by 2 and substituting the value of b in the above two equations we have the value of a as 1 by 4 and the value of c as 3 by 4. We can write x square plus x plus 1 upon x square into x plus 2 is equal to 1 upon 4x plus 1 upon 2x square plus 3 upon 4 into x plus 2. Now integrating both the sides we have integral of x square plus x plus 1 upon x square x plus 2 dx is equal to 1 by 4 integral of dx by x plus 1 by 2 integral of dx upon x square plus 3 by 4 integral of dx upon x plus 2 that is further written as 1 upon 4 log of mod x plus integral of dx upon x square is minus 1 upon 2x then we have plus 3 by 4 log mod of x plus 2 plus c. So we can write the answer and that is first of all 3 upon 4 log mod of x plus 2 plus 1 upon 4 log of mod x minus 1 upon 2x plus c. Right, so this is the required answer to the session. Hope you understood the whole concept well and have a very nice day ahead.