How do you evaluate the integral #int3e^(x)-5e^(2x) dx#?

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Answer 1 · 2014-09-25T20:12:47

#int 3e^x-5e^{2x}dx=3e^x-5/2e^{2x}+C#

Let us look at some details.

Remember:

#int e^x dx=e^x+C#

#int e^{kx} dx=e^{kx}/k+C#

Now, let us work on the integral.

#int 3e^x-5e^{2x}dx#

by applying integral on each term,

#=int 3e^x dx -int 5e^{2x}dx#

by pulling constants out of the integrals

#=3int e^x dx-5int e^{2x}dx#

by applying the formulas above,

#=3e^x-5e^{2x}/2+C#