 Hello and welcome to the session I am the picker here. Let's discuss a question which says evaluate the calling 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. So let's start the solution. t is equal to 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 3 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 i1 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 is equal to now for x square plus 4 is equal to t 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 this is equal to this implies t is equal to so i1 is equal to integral from 4 to 8 log of mod defining its value. So this is equal to log 8 minus log 4 now log m minus log n is equal to log 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 t dx is equal to 1 by 8 tan inverse. So this is equal to now we will find its value from 0 to 2 is equal to 1 over tan this is equal to 1 by 2 1 minus tan this is again equal to 1 over 2 because tan pi by 4 is equal to 1. So this is equal to 1 by 2 into pi by 4 so this is equal to pi by 8 from 0 to now i1 is log 2 so this is equal to the further above question is I hope the solution is clear.