A problematic integral

If the top is a derivative of the bottom in this case:-

[tex]\frac{e^{2x}-1}{e^{2x}+3}[/tex]

derivative of [tex]\frac{d}{dx}e^{2x}-1=2e^{2x}[/tex]

using the log rule for integrals.

[tex]h(x)=\frac{derivative of bottom}{bottom}=ln(bottom)[/tex]

Since it is a derivative by a factor of

[tex](e^{2x}+3)^{2/3}=ln(e^{2x}+3)^{2/3}[/tex]

I think that's what they've done for d/dt.

[tex]\frac {2(e^{2x})}{e^{2x}+3}[/tex]

 

I think your right looking at the answer in d/dx it looks like this

[tex]\frac {2}{3}ln(3+e^{2x})-\frac{1}{6}ln(e^{2x})[/tex]

A bit different.

and for [tex]\frac{d}{dt}[/tex] the paremetric, I get nothing like your books answer either? Sorry I couldn't be of more help, I don't know what they've done then? I'm sure one of the resident experts can figure it out?

[tex] \frac{e^{2x}+1}{e^{2x}-3}=\frac{d}{dx} of bottom=2e^{2x}[/tex]

which is

[tex]\frac {1}{2}(e^{2x})[/tex]

or [tex]\frac{1}{2}(top)[/tex]

was what I was thinking, but this is apparently not right.

 

try solving with the inverse tangent function.

1/[3((e^t/srt3)^2+1)] , try integrating that and tell me if you get what you are looking for

 

It is perhaps simplest to use the variable change [itex]u=e^{x}\to{du}=\frac{du}{u}[/itex]

Then we get:
[tex]\int\frac{e^{2x}-1}{e^{2x}+3}dx=\int\frac{u^{2}-1}{(u^{2}+3)u}du[/tex]
We then use partial fractions decomposition:
[tex]\frac{u^{2}-1}{(u^{2}+3)u}=\frac{Au+B}{u^{2}+3}+\frac{C}{u}\to{C}=-\frac{1}{3}, A=\frac{4}{3},B=0[/tex]
Then your problem is readily solved

 

Arildno did fine. He did not solve the problem in full; he left the final steps up to the original poster.

The only problem with Arildno's post is the use of the word "simplest":

It is even simpler to use the variable change
[tex]u=e^{2x}\to{dx}=\frac1 {2u}du}[/tex]

Then
[tex]\int\frac{e^{2x}-1}{e^{2x}+3}dx\to
\int\frac1 2\; \frac{u-1}{u(u+3)}du[/tex]
Decomposing,
[tex]\frac1 2\frac{u-1}{u(u+3)} = \frac1 6\left(\frac4{u+3}-\frac1u\right)[/tex]
which leads to the easily integrable function
[tex]\int\frac1 6\left(\frac4{u+3}-\frac1u\right)du[/tex]

 

wrong, on the demominator, you cannot factor out u from 3, a constant. look carefully...

 

Which line do you think is wrong?

 

What are you talking about? The technique is called "partial fraction decomposition". A reference: http://mathworld.wolfram.com/PartialFractionDecomposition.html" [Broken]

It is easy to verify that
[tex]\frac1 2\frac{u-1}{u(u+3)} = \frac1 6\left(\frac4{u+3}-\frac1u\right)[/tex]
The least common denominator of the sum on the left hand side is [itex]u(u+3)[/itex]. Expanding the left hand side,
[tex]\frac1 6\left(\frac4{u+3}-\frac1u\right) =
\frac1 6\left(\frac4{u+3}-\frac1u\right)\frac{u(u+3)}{u(u+3)} =
\frac1 6\;\frac{4u-(u+3)}{u(u+3)} = \frac1 6\;\frac{3u-3}{u(u+3)} = \frac1 2\frac{u-1}{u(u+3)}[/tex]
which is of course the right-hand side.

 

what I am trying to say is that the substitution does not equate to the original equation if you observe the denominator, it is not hard to spot, very obvious blunder...

 

Point out this "obvious blunder". The only "obvious blunder" is that you apparently do not know how to use substitution to make integration easier.

Since the OP is long gone, I will finish the integration in post #10.
[tex]\int\frac1 6\left(\frac4{u+3}-\frac1u\right)du =
\frac 1 6\left(4\log(u+3)-\log(u)\right) + C[/tex]
This result is in terms of [itex]u[/itex]. The desired result is in terms of the original variable [itex]x[/itex]. Applying the substitution [itex]u=e^{2x}[/itex],
[tex]\int\frac{e^{2x}-1}{e^{2x}+3}dx =
\frac 1 6\left(4\log(e^{2x}+3)-2x\right) + C =
\frac 1 3\left(2\log(e^{2x}+3)-x\right) + C[/tex]

To verify, differentiate the right-hand side wrt [itex]x[/itex]:
[tex]\frac d{dx} \left(\frac 1 3\left(2\log(e^{2x}+3)-x\right)+C\right) =
\frac 1 3\left(2\frac{2e^{2x}}{e^{2x}+3} - 1\right)[/tex]
Simplifying,
[tex]\frac 1 3\left(2\frac{2e^{2x}}{e^{2x}+3} - 1\right) =
\frac 1 3\;\frac{4e^{2x}-(e^{2x}+3)}{e^{2x}+3} =
\frac{e^{2x}-1}{e^{2x}+3}[/tex]
which of course is the function to be integrated.

 

We have, by the values for A, B, C:
[tex]\frac{\frac{4}{3}u+0}{u^{2}+3}+\frac{\frac{-1}{3}}{u}=\frac{(\frac{4}{3}u+0)u+(-\frac{1}{3})(u^{2}+3)}{(u^{2}+3)u}=\frac{(\frac{4}{3}-\frac{1}{3})u^{2}+0u+(-\frac{1}{3})*3}{(u^{2}+3)u}=\frac{u^{2}-1}{(u^{2}+3)u}[/tex]
as we should have.

In the future, mathgician, please remain silent when you haven't got the competence to make an informed contribution

And, DH:
I DID qualify "simplest" by the word "perhaps"..

 

Here's my treat

[tex] \int \frac{e^{2x}-1}{e^{2x}+3}{}dx =\int \frac{e^{2x}+3-4}{e^{2x}+3}{}dx =x-4\int \frac{e^{-2x}}{1+3e^{-2x}}{}{dx}=x+\frac{2}{3}\int\frac{d\left(1+3e^{-2x}\right)}{1+3e^{-2x}}[/tex]

[tex] =x+\ln\left(\left(1+3e^{-2x}\right)^{\frac{2}{3}}\right) +C [/tex]

Simpler or simplest ?

 

Straightforward, but not necessarily simplest.