- #1
- 1,395
- 0
if what you say is true than the
question must be constructed otherwise
because the derivative of the bottom is NOT what is on TOP??
i saparated them into 2 integrals the second part is the
problematic
because when
i take the whole
1/(e^2x+3) and i and take this as a complex function LAN (as shown in the file)
ln(e^2x+3)/(2*e^2x) it doesnt come out as in the book
what is the law of transforming into lan fuction exept of the case 1/x
i meen in the case of complex function
when i have 1/(something else then simple lenear stuff)????/
arildno Your solution is incorect.
It is perhaps simplest to use the variable change [itex]u=e^{x}\to{du}=\frac{du}{u}[/itex]
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...
wrong, on the demominator, you cannot factor out u from 3, a constant. look carefully...
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...
It is perhaps simplest to use the variable change [itex]u=e^{x}\to{dx}=\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