- #1
transgalactic said: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 doesn't 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)?/
Mathgician said:arildno Your solution is incorect.
arildno said:It is perhaps simplest to use the variable change [itex]u=e^{x}\to{du}=\frac{du}{u}[/itex]
D H said: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]
Mathgician said:wrong, on the demominator, you cannot factor out u from 3, a constant. look carefully...
Mathgician said:wrong, on the demominator, you cannot factor out u from 3, a constant. look carefully...
D H said:What are you talking about? The technique is called "partial fraction decomposition". A reference: http://mathworld.wolfram.com/PartialFractionDecomposition.html"
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.
Mathgician said: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...
arildno said: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
An integral is a mathematical concept that represents the area under a curve on a graph. It is important to be able to solve integrals because they are used in many real-world applications, such as calculating volumes, finding average values, and solving differential equations.
Some common techniques for solving problematic integrals include substitution, integration by parts, partial fractions, and trigonometric substitutions. It is also helpful to have a good understanding of basic integration rules and properties.
Choosing the appropriate technique for a specific integral can be a trial-and-error process. It is often helpful to look for patterns or familiar forms in the integral, and then use the appropriate technique to transform it into a solvable form.
One tip for approaching difficult integrals is to break them down into smaller, more manageable pieces. This can be done through techniques like partial fractions or integration by parts. It is also important to carefully check for any algebraic mistakes or simplifications that can be made before attempting to integrate.
Yes, there are many online resources, such as video tutorials and practice problems, that can provide guidance and help with solving problematic integrals. Additionally, consulting with a math tutor or professor can also be beneficial in understanding and solving difficult integrals.