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Did I do this integral right?

1,197
1
[tex]\int \frac {e^x+4}{e^x}dx =?[/tex]

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

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

Answers and Replies

Galileo
Science Advisor
Homework Helper
1,989
6
Rewriting [itex]\frac{e^x+4}{e^x}=1+4e^{-x}[/itex] was correct.

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

Mind the difference between [itex]x^a[/itex] where the base is the variable and [itex]a^x[/itex] where the base is constant and the exponent is the variable.
 
1,197
1
since [tex]\int e^{x} = e^{x}+C[/tex]
then...
[tex]\int 1+4e^{-x} = x+4e^{-x}+C[/tex]

is that correct?
 
786
0
When you differentiate [tex]x + 4e^{-x} + C[/tex] you get

[tex] 1 - 4e^{-x} [/tex] , so the integral is actually

[tex] \int 1 + 4e^{-x} dx = x - 4e^{-x} + C [/tex]
 
1,197
1
I dont understand where the negative came from
 
786
0
The integral of [tex] e^x dx = {e^x} + C[/tex]

The integral of [tex] e^{-x}dx = -e^{-x} + C[/tex].

Differentiating that answer you find that [tex] \frac {d}{dx} -e ^{-x} = e^{-x}
[/tex]
 
dextercioby
Science Advisor
Homework Helper
Insights Author
12,965
536
Think of it as an [itex] e^{u} [/itex] and apply the method of substitution:
[tex] -x=u [/tex]

Daniel.

P.S.That's how u end up with the minus.
 
1,197
1
[tex]\int \frac {e^x}{e^x+4}dx =?[/tex]

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

Did I do this correctly? Is there a more simplified answer?
 
NateTG
Science Advisor
Homework Helper
2,448
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UrbanXrisis said:
Did I do this correctly? Is there a more simplified answer?
Looks good to me.
 
dextercioby
Science Advisor
Homework Helper
Insights Author
12,965
536
Don't forget the constant of integration.

Daniel.
 

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