How to integrate? #int 2e^x+2e^-x dx#

1 Answer
jim p.
Nov 28, 2017

# 2(e^x - e^(-x)) + c#

Explanation:

I don't see the "dx" in there, I'm going to assume your problem as given to you has it.

#int(2e^x + 2e^(-x))dx = 2inte^xdx + 2inte^(-x)dx#

...for the first term, you know that the antiderivative of #e^x = e^x#

for the second term, remember that #int(e^u du) = e^u#. In this case, #u = -x#, so du = -1dx. So you can convert the second term to a form you can readily integrate by multiplying by #(-1)(-1)#, which turns your second integral term into:

#-2inte^(-x) (-dx)#

so, your original problem's integral evaluates to:

#2e^x - 2e^(-x) + c#

#= 2(e^x - e^(-x)) + c#

GOOD LUCK

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