 Hello, I am welcome to the session, I am Deepika here, let's discuss the question which says evaluate the following integral using substitution integral from 0 to 2 dx upon x plus 4 minus x square. So, let's start the solution. Given integral is integral from 0 to 2 dx upon x plus 4 minus x square. Now, this can be written in the form integral from 0 to 2 dx upon minus x square minus x minus 4, we have taken minus common and this is again equal to minus integral from 0 to 2 dx upon x square minus x minus 4, now let us subtract minus 1 by 4 and add 1 by 4. Now this is again equal to minus integral from 0 to 2 dx upon x square plus 1 by 4. Minus x minus 4 minus 1 by 4 and this is equal to minus integral from 0 to 2 dx upon x minus 1 by 2 whole square minus 17 over 4 and this is equal to minus integral from 0 to 2 dx upon x minus 1 by 2 whole square minus under root of 17 over 2 square and this is equal to integral from 0 to 2 dx upon under root of 17 over 2 square minus x minus 1 by 2 whole square. Now here I put x minus 1 by 2 is equal to t then dx is equal to dt, so the new limits are when x is equal to 0 this implies t is equal to minus 1 over 2 and when x is equal to 2 this implies t is equal to 3 over 2, so therefore integral from 0 to 2 dx upon x plus 4 minus x square is equal to integral from minus 1 by 2 to 3 by 2 dt upon under root of 17 by 2 square minus t square. Now we know that integral of dx upon a square minus x square is equal to 1 over 2a log of mod a plus x over a minus x plus c, so this integral is equal to 1 over 2a. Now a is under root of 17 over 2, so 2 into under root of 17 over 2 into log of mod under root of 17 over 2 plus t upon under root of 17 over 2 minus t from minus 1 by 2 to 3 by 2 and this is equal to 1 over under root of 17 log of mod under root of 17 plus 2t upon under root of 17 minus 2t from minus 1 by 2 to 3 by 2. Now we will substitute the limits, so this is equal to 1 over under root of 17 log of mod under root of 17 plus 2 into 3 by 2 over under root of 17 minus 2 into 3 by 2 minus log mod of under root of 17 plus 2 into minus 1 by 2 over under root of 17 minus 2 into minus 1 by 2 and this is equal to 1 over under root of 17 log mod of under root of 17 plus 3 upon under root of 17 minus 3 minus log under root of 17 minus 1 over under root of 17 plus 1 and this is again equal to 1 over under root of 17 into log. Now let us multiply the numerator and denominator by under root of 17 plus 3, so we have under root of 17 plus 3 whole square over under root of 17 square minus 3 square minus log here multiply the numerator and the denominator by root 17 minus 1, so we have under root of 17 minus 1 whole square upon under root of 17 square minus 1 square and this is equal to 1 over under root of 17 log of mod, now here apply the formula of a plus b whole square, so we have 17 plus 9 plus 6 root 17 over 17 minus 9 minus log mod of 17 plus 1 minus 2 root 17 over 17 minus 1, so this is equal to 1 over under root of 17 log mod 26 plus 6 root 17 over 8 minus log mod of 18 minus 2 root 17 over 16, now log m minus log n is log m over n, so this is equal to 1 over under root of 17 log mod of 26 plus 6 root 17 over 8 into 16 minus 2 root 17, so this is equal to 1 over under root of 17 log mod of 2 into 26 plus 6 root 17 over 18 minus 2 under root of 17, now let us take 2 common from the numerator as well as from the denominator, we have this is equal to 1 over under root of 17 log mod of 4 into 13 plus 3 root 17 over 2 into 9 minus under root of 17, so this is equal to 1 over under root of 17 log mod of 2 into 13 plus 3 under root of 17 over 9 minus under root of 17 and this is equal to 1 over under root of 17 log mod of 2 into 13 plus 3 under root of 17 over 9 minus under root of 17 into, let us multiply the numerator as well as the denominator by 9 plus under root of 17, so this is equal to 1 over under root of 17 log mod of 2 into 117 plus 13 under root of 17 plus 27 under root of 17 plus 51 upon 9 square minus 17, so this is again equal to 1 over under root of 17 log mod of 2 into 168 plus 40 under root of 17 over 64 and this is equal to 1 over under root of 17 log mod of 2 into let us take 8 common from this bracket, so this is 21 plus 5 under root of 17 over 64 and this is equal to 1 over under root of 17 log mod of 21 plus 5 under root of 17 upon 4, hence the answer for the above question is 1 over under root of 17 log 21 plus 5 under root of 17 over 4. I hope the solution is clear to you, bye and take care.