Is My Solution to \int \frac {e^x+4}{e^x}dx Correct?

[UrbanXrisis]

\int \frac {e^x+4}{e^x}dx =?

Here's what I did:
\int \frac {e^x+4}{e^x}dx = \int e^{-x}(e^x+4)dx
\int e^{-x}(e^x+4)dx =\int 1+4e^{-x} = x-\frac {4e^{-x+1}}{x+1}

Did I do this correctly? Is there a more simplified answer?

 

[Galileo]

Rewriting \frac{e^x+4}{e^x}=1+4e^{-x} was correct.

Check the antiderivative of e^{-x}. Your answer is not correct. You can easily check it by differentiating it.

Mind the difference between x^a where the base is the variable and a^x where the base is constant and the exponent is the variable.

 

[UrbanXrisis]

since \int e^{x} = e^{x}+C
then...
\int 1+4e^{-x} = x+4e^{-x}+C

is that correct?

 

[Jameson]

When you differentiate x + 4e^{-x} + C you get

1 - 4e^{-x} , so the integral is actually

\int 1 + 4e^{-x} dx = x - 4e^{-x} + C

 

[UrbanXrisis]

I don't understand where the negative came from

 

[Jameson]

The integral of e^x dx = {e^x} + C

The integral of e^{-x}dx = -e^{-x} + C.

Differentiating that answer you find that \frac {d}{dx} -e ^{-x} = e^{-x}<br />

 

[dextercioby]

Think of it as an e^{u} and apply the method of substitution:
-x=u

Daniel.

P.S.That's how u end up with the minus.

 

[UrbanXrisis]

\int \frac {e^x}{e^x+4}dx =?

Here's what I did:
= \int e^{x}(e^x+4)^{-1}dx
subsitute:
u=e^x+4
du=e^x dx
\int u^{-1}du =ln(e^x+4)

Did I do this correctly? Is there a more simplified answer?

 

[NateTG]

Did I do this correctly? Is there a more simplified answer?

Looks good to me.

 

[dextercioby]

Don't forget the constant of integration.

Daniel.