Solving a Problematic Integral: Tips and Tricks | Help with Difficult Integrals

• transgalactic
In summary, Arildno did not correctly solve the problem. He left the final steps up to the original poster.
transgalactic
i added a file with with this integral

in which i tried to solve it

how do i solve this case?

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• integral.jpg
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If the top is a derivative of the bottom in this case:-

$$\frac{e^{2x}-1}{e^{2x}+3}$$

derivative of $$\frac{d}{dx}e^{2x}-1=2e^{2x}$$

using the log rule for integrals.

$$h(x)=\frac{derivative of bottom}{bottom}=ln(bottom)$$

Since it is a derivative by a factor of

$$(e^{2x}+3)^{2/3}=ln(e^{2x}+3)^{2/3}$$

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

$$\frac {2(e^{2x})}{e^{2x}+3}$$

Last edited:
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)?/

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)?/

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

$$\frac {2}{3}ln(3+e^{2x})-\frac{1}{6}ln(e^{2x})$$

A bit different.

and for $$\frac{d}{dt}$$ 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?

$$\frac{e^{2x}+1}{e^{2x}-3}=\frac{d}{dx} of bottom=2e^{2x}$$

which is

$$\frac {1}{2}(e^{2x})$$

or $$\frac{1}{2}(top)$$

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

Last edited:
how i deside if its this is lan function

i can take every time all the function on the buttom
put it in a lan
an divide it in its derivative?

even if i got a fuction
1/(x+2)^2

its integral solves [(x+2)^-1]/-1

or i can open this equetion (a+b)^2=x^2+2*x*2+2^2
then put it all in a lan ln(x^2+2*x*2+2^2) and i will divide it by it derivative

2x+4

ln(x^2+2*x*2+2^2)/2x+4

what i need to do?
can some one make reason in this problem

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 $u=e^{x}\to{du}=\frac{du}{u}$

Then we get:
$$\int\frac{e^{2x}-1}{e^{2x}+3}dx=\int\frac{u^{2}-1}{(u^{2}+3)u}du$$
We then use partial fractions decomposition:
$$\frac{u^{2}-1}{(u^{2}+3)u}=\frac{Au+B}{u^{2}+3}+\frac{C}{u}\to{C}=-\frac{1}{3}, A=\frac{4}{3},B=0$$

thank u very much

Mathgician 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":
arildno said:
It is perhaps simplest to use the variable change $u=e^{x}\to{du}=\frac{du}{u}$

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

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

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
$$u=e^{2x}\to{dx}=\frac1 {2u}du}$$

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

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...

Which line do you think is wrong?

Mathgician said:
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"

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

Last edited by a moderator:
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
$$\frac1 2\frac{u-1}{u(u+3)} = \frac1 6\left(\frac4{u+3}-\frac1u\right)$$
The least common denominator of the sum on the left hand side is $u(u+3)$. Expanding the left hand side,
$$\frac1 6\left(\frac4{u+3}-\frac1u\right) = \frac1 6\left(\frac4{u+3}-\frac1u\right)\frac{u(u+3)}{u(u+3)} = \frac1 6\;\frac{4u-(u+3)}{u(u+3)} = \frac1 6\;\frac{3u-3}{u(u+3)} = \frac1 2\frac{u-1}{u(u+3)}$$
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...

Last edited by a moderator:
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...

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.
$$\int\frac1 6\left(\frac4{u+3}-\frac1u\right)du = \frac 1 6\left(4\log(u+3)-\log(u)\right) + C$$
This result is in terms of $u$. The desired result is in terms of the original variable $x$. Applying the substitution $u=e^{2x}$,
$$\int\frac{e^{2x}-1}{e^{2x}+3}dx = \frac 1 6\left(4\log(e^{2x}+3)-2x\right) + C = \frac 1 3\left(2\log(e^{2x}+3)-x\right) + C$$

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

arildno said:
It is perhaps simplest to use the variable change $u=e^{x}\to{dx}=\frac{du}{u}$

Then we get:
$$\int\frac{e^{2x}-1}{e^{2x}+3}dx=\int\frac{u^{2}-1}{(u^{2}+3)u}du$$
We then use partial fractions decomposition:
$$\frac{u^{2}-1}{(u^{2}+3)u}=\frac{Au+B}{u^{2}+3}+\frac{C}{u}\to{C}=-\frac{1}{3}, A=\frac{4}{3},B=0$$

We have, by the values for A, B, C:
$$\frac{\frac{4}{3}u+0}{u^{2}+3}+\frac{\frac{-1}{3}}{u}=\frac{(\frac{4}{3}u+0)u+(-\frac{1}{3})(u^{2}+3)}{(u^{2}+3)u}=\frac{(\frac{4}{3}-\frac{1}{3})u^{2}+0u+(-\frac{1}{3})*3}{(u^{2}+3)u}=\frac{u^{2}-1}{(u^{2}+3)u}$$
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"..

Last edited:
Here's my treat

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

$$=x+\ln\left(\left(1+3e^{-2x}\right)^{\frac{2}{3}}\right) +C$$

Simpler or simplest ?

Straightforward, but not necessarily simplest.

1. What is an integral and why is it important to be able to solve it?

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.

2. What are some common techniques for solving problematic integrals?

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.

3. How can I determine which technique to use for a specific integral?

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.

4. What are some tips for approaching difficult integrals?

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.

5. Are there any resources available for help with solving problematic integrals?

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.

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