Integration Help

[MrCodeDude]

How do I integrate

( e^4x - e^2x + 1 ) / e^x

Thanks in advance!

 

[virtueixi]

I'll give you a hint. ln ^e ^x= x

 

[MrCodeDude]

But can I just ln only one side?

 

[eakers]

do you know how to integrate using substitution?
edit: that might not help though now that i look better.

 

[MrCodeDude]

NM this response 😉

 

[virtueixi]

ln(e^4x - e^2x + 1)/ln( e^x)

 

[eakers]

hehe yeah just minus one from all the e exponets and integrate

 

[MrCodeDude]

Originally posted by: virtueixi
ln(e^4x - e^2x + 1)/ln( e^x)

Wouldn't you have to ln both sides though?

 

[PowerMacG5]

First seperate it into the sum of integrals.
So you have (after dividing, and changing 1/e^x to e^(-x))
int(e^(3x))dx - int(e^x)dx + int(e^(-x))dx

Now it's simple substitutions. For the first one:
int(e^(3x))dx
let u=3x so (1/3)du = dx
so you have (1/3)int(e^u)du which is equal to (1/3)*e^(3x) or (e^(3x))/3

next you have:
int(e^x)dx which is simply e^x

and lastly you have:
int(e^(-x))dx
let u = -x
-du = dx
so it is -int(e^u)du which is -e^(-x)

So now to put it back together as
((e^(3x))/3) - (e^x) - (e^(-x))

Now if you want to simplify it further you can look at this. The definition of the hyperbolic cosine funcion is (e^x-e^(-x))/2 so ((e^x) - (e^(-x))) is equal to two times the hyperbolic cosing function.
So in the simplest form:

((e^(3x))/3) - 2cosh(x)

 

[PowerMacG5]

Originally posted by: virtueixi
ln(e^4x - e^2x + 1)/ln( e^x)

Look at my answer, you are overanalyzing. As my teacher says, don't look to do more work than is needed. My answer is the simplest, and most efficient way to do it (seperate into the sum of integrals rather than an integral of sums).

 

[MrCodeDude]

((e^(3x))/3) - (e^x) - (e^(-x))

Is that more appropriate than this:

e^4x - 3e^2x - 3
3e^x

 

[PowerMacG5]

Originally posted by: MrCodeDude
((e^(3x))/3) - (e^x) - (e^(-x))

Is that more appropriate than this:

e^4x - 3e^2x - 3
3e^x

Look at my answer/explanation.