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 $$\frac {d}{dx} -e ^{-x} = e^{-x}$$

 

[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.