 Hi, and welcome to our session where we discuss the following question. The question says integrate the following functions given function is x square plus x plus 1 by x plus 1 whole square into x plus 2. Let's now begin with the solution. In this question we have to integrate the function x square plus x plus 1 by x plus 1 whole square into x plus 2 with respect to x. Now this is a proper rational function so we are going to first resolve it into partial fractions. Let x square plus x plus 1 by x plus 1 whole square into x plus 2 is equal to a by x plus 1 plus b by x plus 1 whole square plus c by x plus 2. Now this implies x square plus x plus 1 is equal to a into x plus 1 into x plus 2 plus b into x plus 2 plus c into x plus 1 whole square. Now put x equals to minus 2 so we have 4 minus 2 plus 1 equals to c into minus 2 plus 1 whole square. Now this implies c is equal to 3. Now we are going to put x as minus 1 so we have 1 minus 1 plus 1 equals to b into minus 1 plus 2. This implies b is equal to 1. Now we will put x as 0 so we have 1 equals to a into 1 into 2 plus b into 2 plus c into 1. Now this implies 1 is equal to 2 a plus 2 b plus c. This implies 1 is equal to 2 a plus 2 plus 3. This implies minus 4 is equal to 2 a and this implies a is equal to minus 2 so x square plus x plus 1 by x plus 1 whole square into x plus 2 is equal to minus 2 by x plus 1 plus 1 by x plus 1 whole square plus 3 into x plus 2. Now this implies integral of x square plus x plus 1 by x plus 1 whole square into x plus 2 with respect to x is equal to minus 2 into integral 1 by x plus 1 dx plus integral 1 by x plus 1 whole square dx plus 3 into integral 1 by x plus 2 dx. We know that integral of 1 by ax plus b with respect to x is equal to law mod ax plus b divided by derivative of ax plus b that is a plus c so using this integral of 1 by x plus 1 with respect to x is log mod x plus 1 divided by derivative of x plus 1 that is 1 plus integral now this can be written as x plus 1 to the power minus 2 dx plus 3 integral of 1 by x plus 2 with respect to x is log mod x plus 2 divided by derivative of x plus 2 that is 1 plus c now here c is denoted the sum of all the constant of these integrals now this is equal to minus 2 log mod x plus 1 plus we know that integral of ax plus b to the power n with respect to x is ax plus b to the power n plus 1 by n plus 1 into derivative of ax plus b that is a plus c so using this this integral is equal to x plus 1 to the power minus 1 by minus 1 plus 3 into log mod x plus 2 plus c and this is equal to minus 2 log mod x plus 1 minus 1 by x plus 1 plus 3 log mod x plus 2 plus c hence our required answer is minus 2 log mod x plus 1 minus 1 by x plus 1 plus 3 into log mod x plus 2 plus c so this completes the session bye and take care