How to integrate #int(e^2x+e^-2x)/(e^2x-e^-2x)dx?
3 Answers
We can rewrite as
#I = int (e^(2x) + 1/e^(2x))/(e^(2x) - 1/e^(2x)) dx#
#I = int((e^(4x) + 1)/e^(2x))/((e^(4x) - 1)/e^(2x)) dx#
#I = int(e^(4x) + 1)/(e^(4x) - 1)dx#
Let
#I = 1/4 int (e^u + 1)/(e^u - 1)du#
Now use partial fractions.
#A/(e^u - 1) + B/1 = (e^u + 1)/(e^u - 1)#
#A + Be^u - B = e^u + 1#
We can readily see that
Thus
#I = 1/4(int 2/(e^u - 1) + 1 du)#
#I = 1/2ln|e^u - 1| + 1/4u + C#
#I = 1/2ln|e^(4x) - 1| + x + C#
Hopefully this helps!
The answer is
Explanation:
The function is
Therefore, the integral is
Let
So, the integral is
We have,
Let's Substitute
So,
So,
Hope this helps.