Home > Calculus > Integrals of Exponential Functions

How do you integrate #int (5-e^x)/(e^(2x))dx#?

Answer 1

Paisley Johnson

#-5/2e^(-2x)+e^-x+C#

rearrange as folows.

#int((5-e^x)/e^(2x))dx#

#=int(5/e^(2x)-e^x/e^(2x))dx#

#int(5e^(-2x)-e^(-x))dx#

#=-5/2e^(-2x)+e^-x+C#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Scarlett Allison

To integrate ( \frac{{5 - e^x}}{{e^{2x}}} ) with respect to ( x ), you can use substitution method. Let ( u = e^x ), then ( du = e^x dx ).

Now, substitute ( u = e^x ) and ( du = e^x dx ) into the integral:

[ \int \frac{{5 - e^x}}{{e^{2x}}} dx = \int \frac{{5 - u}}{{u^2}} du ]

[ = \int \left( \frac{5}{u^2} - \frac{1}{u} \right) du ]

[ = -\frac{5}{u} - \ln|u| + C ]

Substitute back ( u = e^x ):

[ = -\frac{5}{e^x} - \ln|e^x| + C ]

[ = -\frac{5}{e^x} - x + C ]

So, the integral of ( \frac{{5 - e^x}}{{e^{2x}}} ) with respect to ( x ) is ( -\frac{5}{e^x} - x + C ), where ( C ) is the constant of integration.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.