 Hello and 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 x into under root of x plus 2 dx. So let's start the solution. Plus 2 is equal to t square dx is equal to 2 dx to the new limits r. Now when x is equal to 0, this implies t is equal to root 2 and when x is equal to 2, this implies t is equal to 2. Therefore, integral from 0 to 2 x into under root of x plus 2 dx is equal to integral from root 2 to 2. Now x is t square minus 2, root of x plus 2 is t and dx is 2t dt. So this is equal to 2 integral from root 2 to 2 square into t square minus 2 dt and this is equal to integral from root 2 to 2 t is to power 4 minus 2t square dt and this is equal to into t is to power 5 over 5 minus 2t cube over 3. Now this is equal to into 2 is to power 5 over 5 minus 2 into 2 is to power 3 over 3 minus is to power 5 over 5 minus 2 into root 2 is to power 3 over 3 and this is equal to into 2 is to power 5 is 32 upon 5 minus 16 upon 3 minus 2 upon 5 minus 4 root 2 upon 3 is equal to 2 into now take 15 as Lcm we have 96 minus 18 minus here also take 15 as Lcm 12 root 2 minus 20 root 2 now this is equal to 2 into 2 over 15 plus 8 root 2 over 15 and this is equal to 32 over 15 plus 16 root 2 over 15 take 16 root 2 over 15 common we have this is equal to the answer for that question is 16 upon 15 into root 2 plus 1. I hope the solution is clear to you