 Hello students, let's solve the following question on integration. We have to integrate the function e to the power 2x minus e to the power minus 2x upon e to the power x plus e to the power minus 2x. Let us now proceed on with the solution and let i be the integral e to the power 2x minus e to the power minus 2x upon e to the power 2x plus e to the power minus 2x dx. Now here we see that the derivative of e to the power 2x plus e to the power minus 2x is 2 into e to the power 2x minus e to the power minus 2x which contains the expression in the numerator. So put t equal to e to the power 2x plus e to the power minus 2x. So dt by dx is equal to 2e to the power 2x minus 2e to the power minus 2x. So this implies dt is equal to 2e to the power 2x minus e to the power minus 2x dx. And this again implies e to the power 2x minus e to the power minus 2x dx is equal to dt by 2. Now e to the power 2x minus e to the power minus 2x dx is equal to dt by 2 and e to the power 2x plus e to the power minus 2x is t. So substituting all these values in this integral the integral i becomes 1 upon t dt by 2 which is again equal to 1 by 2 integral 1 upon t dt. Now we know that the integral of 1 upon t is log t plus c. So this is equal to 1 by 2 log plus c. Now substitute the value of t becomes 1 by 2 log e to the power 2x plus e to the power minus 2x plus c. Hence the integral of the given function is 1 by 2 log e to the power 2x plus e to the power minus 2x plus c. And this completes the question and the session. Bye for now. Take care. Have a good day.