 Hello and welcome to the session let us discuss the following question it says integrate the following function the given function is e to the power tan inverse x upon 1 plus x square. Let us now move on to the solution and let i be the integral e to the power tan inverse x upon 1 plus x square dx. Now here we see that the derivative of tan inverse x is 1 upon 1 plus x square that is this thing. So we put equal to tan inverse x. So dt by dx is equal to 1 upon 1 plus x square and this implies dt is equal to 1 upon 1 plus x square dx. Now 1 upon 1 plus x square into dx is dt and tan inverse x is t. So the integral i becomes integral of e to the power t dt. Now we know that the integral of e to the power t dt is e to the power t plus c. So this is equal to e to the power t plus c where c is the constant of the integration. Now substitute the value of t our t is tan inverse x plus c. Hence the integral of the given function is e to the power tan inverse x plus c and this completes the question. Bye for now. Take care. Have a good day.