 Hello and welcome to the session I am Deepika here. Let's discuss a question which says evaluate the following definite integral integral from 0 to 2 6x plus 3 upon x square plus 4 dx. Now in this question we will use a second fundamental theorem of integral calculus. Let's start the solution. Let i is equal to given integral integral from 0 to 2 6x plus 3 upon x square plus 4 dx. Now we can rewrite this integral as integral from 0 to 2 6x upon x square plus 4 dx plus integral from 0 to 2 3 upon x square plus 4 dx and this is again equal to 3 integral of 0 to 2 2x upon x square plus 4 dx plus integral from 0 to 2 3 upon x square plus 4 dx. Let i is equal to integral from 0 to 2 2x upon x square plus 4 dx and i2 is equal to integral from 0 to 2 3 upon x square plus 4 dx. So we have i is equal to 3 i1 plus i2. Now to solve the given integral we will solve these two integrals. So let us consider i1. Now integral from 0 to 2 2x upon x square plus 4 dx is equal to now put x square plus 4 is equal to t then 2x dx is equal to dt and x dx is equal to 1 by 2 dt. So the new limits are when x is equal to 0 this implies p is equal to 4 and when x is equal to 2 this implies t is equal to 8. So i1 is equal to integral from 4 to 8 upon t and this is equal to log of log t. Now we will find its value from 4 to 8. So this is equal to log 8 minus log 4. Now log m minus log n is equal to log m over n. This is equal to log 8 over 4 which is equal to log 2. Now we will solve i2 which is equal to integral from 0 to 2 1 over x square plus 4 dx and we can rewrite this integral as integral from 0 to 2 1 over x square plus 2 square dx. Now integral of 1 over x square plus a square dx is equal to 1 by 8 tan inverse x over 8 plus c. So this is equal to 1 over 2 tan inverse x by 2. Now we will find its value from 0 to 2. So this is equal to 1 over 2 into tan inverse 2 by 2 minus tan inverse 0 by 2 and this is equal to 1 by 2 tan inverse 1 minus tan inverse 0 and this is again equal to 1 over 2. Now tan inverse 1 is equal to pi by 4 because tan pi by 4 is equal to 1. So this is equal to 1 by 2 into pi by 4 minus 1 by 2 into 0 which is 0. So this is equal to pi by 8. Therefore i which is equal to integral from 0 to 2 6x plus 3 upon x square plus 4 dx is equal to 3 into i1 plus i2 and this is equal to 3 into now i1 is log 2 plus i2 is pi by 8. So this is equal to 3 log 2 plus 3 pi by 8. Hence the answer for the above question is 3 log 2 plus 3 pi by 8. I hope the solution is clear to you. Bye and take care.