 Hello and welcome to the session I am Deepika here. Let's discuss the question which says evaluate the following definite integral integral from 1 to 2, 5x square upon x square plus 4x plus 3 dx. Now in this question we will use a second fundamental theorem of integral calculus. So let's start the solution. Let i is equal to integral from 1 to 2, 5x square upon x square plus 4x plus 3 dx. Now we will first solve the indefinite integral 5 into integral of x square upon x square plus 4x plus 3 dx. Now we will integrate this rational function by partial fractions. Now here the integral x square upon x square plus 4x plus 3 is not a proportional function. So we will divide that plus 4 therefore the integral x square x square plus 4x plus 3 is equal to 1 is equal to 1 which determines x plus 3 plus b into x plus 1 to ax plus 3a the coefficients of the constant term. We have 4 is equal to equation as number 1 and this as number 2. Now we will solve these equations is equal to minus 2 which is equal to 9 by 2. Here plus 4x plus 3 is here plus 4x plus 3 1 by 2 mod x plus 3 that is the anti derivative of the given function 5x square upon x square plus 4x plus 3 is equal to 1 by 2 log 1 which is 5 into 1 plus 1 by 2 log 1 plus 1 to minus 9 by 2 log 1 plus 3 4 45 over 2 log 4 this is equal to into log 5 minus log 4 and this log x because log 2 log m minus log n into common from these 2 is log 3 over 2 and log 5 by 4. I hope the solution is clear to you. Bye and take care.