 Hi and welcome to the session. I am Purva and I will help you with the following question. Choose the correct answer. Integral dx upon e raised to the power x plus e raised to the power minus x is equal to a tan inverse e raised to the power x plus c b tan inverse e raised to the power minus x plus c c log e raised to the power x minus e raised to the power minus x plus c and d log e raised to the power x plus e raised to the power minus x plus c. Now we begin with the solution. Let us denote i by integral dx by e raised to the power x plus e raised to the power minus x So we have I is equal to integral dx upon e raised to the power x plus e raised to the power minus x and we can also write this as this is equal to integral dx upon e raised to the power x plus 1 upon e raised to the power x. is equal to integral dx upon now taking the LCM we get e raised to the power 2x plus 1 upon e raised to the power x and this is further equal to integral e raised to the power x upon e raised to the power x whole square. We can write e raised to the power 2x as e raised to the power x whole square plus 1 dx. Now we put e raised to the power x equal to t. So differentiating we get e raised to the power x dx is equal to dt. Putting these values in i we get i is equal to integral dt upon t square plus 1. Now integrating we get this is equal to now integral 1 upon t square plus 1 is equal to tan inverse t plus c. Now we know that t is equal to e raised to the power x. So putting this value we get this is equal to tan inverse e raised to the power x plus c and this is equal to our option a. So we get our answer as a. Hope you have understood the solution. Take care and God bless you.