 Hello and welcome to the session. The question says integrate the following function and the 27th one as x plus 2 upon root over x square plus 2x plus 3. So we have to integrate this function. Therefore we have integral x plus 2 upon root over x square plus 2x plus 3 into dx. Now let us take x plus 2 is equal to a into derivative of the polynomial x square plus 2x plus 3 plus b which further implies that x plus 2 is equal to a into 2x plus 2 plus b. Or we have a is equal to 1 upon 2. Now repeating the coefficients we have 2 is equal to 2a plus b which implies that b is equal to 2 minus 2a and a is 1 upon 2. So this is equal to 1. Thus the integral x plus 2 upon root over x square plus 2x plus 3 can be written as now x plus 2 is a into 2x plus 2 plus b. So a is half, we have half into 2x plus 2 plus b is 1 integral upon the denominator is x square plus 2x plus 3 into dx. This is further equal to half integral 2x plus 2 upon root over x square plus 2x plus 3 into dx plus integral 1 into dx upon root over x square plus 2x plus 3. Now let us solve both of these integrals one by one. The first one is integral 2x plus 2 upon root over x square plus 2x plus 3 into dx. Let us take t is equal to x square plus 2x plus 3 so this implies dt is equal to 2x plus 2 into dx differentiating both sides with respect to x. So this integral can further be written as integral dt upon root over t. That is t raise to the power minus half into dt and this is further equal to 2 root t plus c1 where c1 is a constants integral of x raise to power n with respect to x is equal to x raise to the power n plus 1 upon n plus 1. So by using this formula on integrating this function we get 2 root 2. Now 2 root t let us put the value of t which is x square plus 2x plus 3 plus we have a constant c1. So this is the value of the first integral. Let us write it down here. So this is half into 2 root over x square plus 2x plus 3 plus c1. Now we have to integrate this function. This is integral dx upon root over x square plus 2x plus 3. Now to write a polynomial of the form ax square plus bx plus c as the square of the sum of 2 polynomials formula is a into x plus b upon 2a whole square plus c upon a minus b square upon 4a square. So with the help of this formula let us write the polynomial x square plus 2x plus 3 as the sum of the squares of 2 polynomials. So this is x plus 1 whole square plus minus and this is further equal to x plus 1 whole square minus 1 square. So the given integral can be written as dx upon root over x plus 1 whole square minus 1 square. Now let us take t is equal to x plus 1. So this implies dt is equal to dx. So this integral can further be written as dt upon root over t square minus 1. Now the formula to integrate a function of the type 1 upon root over x square minus a square with respect to x is equal to log mod x plus root over x square minus a square plus a constant c. So on integrating this function we have log mod t plus root over t square minus 1 plus a constant let c2. So this can further be written as log mod t is equal to x plus 1 plus here t square plus 1 is x plus 1 whole square minus 1 square and this is equal to x square plus 2x plus 3 plus a constant c2. Thus let us write down the value of this integral which is log mod x plus 1 plus root over x square plus 2x plus 3 plus a constant c2. So this can further be written as 2 cancels output 2 and we have root over x square plus 2x plus 3 plus log mod x plus 1 plus root over x square plus 2x plus 3 plus c2 which is further equal to root over x square plus 2x plus 3 plus log mod x plus 1 plus root over x square plus 2x plus 3 plus c where c is equal to the sum of constant c1 and c2. Thus on integrating the given function we get root over x square plus 2x plus 3 plus log mod x plus 1 plus root over x square plus 2x plus 3 plus a constant c. So this completes the session. Bye and take care.