Mastering a Tricky Integral: Solving Linear 1st Order ODEs with x, Log, and e

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In summary, the conversation discusses a problem involving an integral that came up while solving a linear 1st order ODE. The participants suggest different approaches, including using u substitution and integration by parts. However, it is mentioned that there may not be a known formula for this integral. Eventually, it is revealed that the correct integrating factor is (e^x + 1) and the resulting integral is much simpler than previously thought. The conversation also reflects on a mistake made in the initial approach to the problem.
  • #1
uart
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The following integral came up in as part of solving a linear 1st order ODE and it's stumped me.

[tex]\int \frac{ x \log ( e^x + 1)}{e^x+1}[/tex]

Does anyone know of a good approach to tackle this one?
 
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  • #2
Maybe a u substituion and integration by parts would work.
Let u = ex + 1
 
  • #3
d_leet said:
Maybe a u substituion and integration by parts would work.
Let u = ex + 1

I agree. e^x, x and log() don't belong together.
However integrating by parts will only lead to longer and more tedious work involving many more substitutions and multiple integration by parts.
But it seems to be the only way.Multiplying num and den by e^x might help.
 
  • #4
There's no formula on this one found in Mathematica's library. Most likely there's no combination of known special functions which would provide you with an antiderivative.

Daniel.
 
  • #5
dextercioby said:
There's no formula on this one found in Mathematica's library. Most likely there's no combination of known special functions which would provide you with an antiderivative.

Daniel.

Yes that's exactly what I previously thought Daniel. But it does have an anit-derivative, I'm just don't know how it was deduced.


The anti-derivative is

y = [tex] (\frac{x^2}{2} + 2) \frac{ \log( e^x + 1)} {e^x + 1}} [/tex]
 
  • #6
uart said:
Yes that's exactly what I previously thought Daniel. But it does have an anit-derivative, I'm just don't know how it was deduced.


The anti-derivative is

y = [tex] (\frac{x^2}{2} + 2) \frac{ \log( e^x + 1)} {e^x + 1}} [/tex]

Well, it works if "log" is "ln" anyway.

-Dan
 
  • #7
uart said:
y = [tex] (\frac{x^2}{2} + 2) \frac{ \log( e^x + 1)} {e^x + 1}} [/tex]
Uhmm, just wonder, but have you tried to differentiate that monster to see if it returns:
[tex]\frac{x \ln(e ^ x + 1)}{e ^ x + 1}[/tex]?
I guess not... :rolleyes:
 
  • #8
Hang on a minute, the above doesn't seem to be correct. Let me post the full problem.

The problem was to solve the following DE (where y' denotes dy/dx),

[tex] y^{\prime} + \left( \frac{e^x} {e^x + 1} \right) y = \frac{x}{e^x + 1} [/tex]

With the initial condition of [tex]y(0) = 1[/tex]
 
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  • #9
DOH! My appologies to everyone, I made a huge stuff up on this problem which lead me to the wrong integral. It's actually much easier than I thought, I'll post more details in one moment when I work it out.
 
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  • #10
OK here's the full story. I was helping my niece with her maths (2nd year uni) and this was a problem she was stuck on. She had worked out the "integrating factor" for this question as [tex]u = \log(e^x + 1)[/tex] (which is wrong), but I checked her working (twice) and thought it was correct. Sorry my stuff up. It was a case of the old "you're checking someones work and follow their same mistakes" thing. As soon as I looked at the problem on my own just now I immediately saw the error.

The integrating factor was actually just [tex]u = (e^x+1)[/tex], which leads to a trivial integral instead of the monster I previously posted.
 
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  • #11
The final integral needed to solve the above DE is,

[tex] u y = \int \frac{u x}{e^x + 1} [/tex]

When I put [tex]u = \log(e^x + 1)[/tex] I got the horrible integral that I originally posted.

Of course with [tex]u = (e^x + 1)[/tex], as it should have been, you get the truly simple integral of,
[tex]\int \, x dx[/tex]

DOH I feel stupid now. :eek:
 
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FAQ: Mastering a Tricky Integral: Solving Linear 1st Order ODEs with x, Log, and e

What is an integral?

An integral is a mathematical concept that represents the area under a curve in a graph. It is also known as the antiderivative of a function.

Why are integrals important?

Integrals are important because they are used in many areas of science, such as physics, engineering, and economics, to solve real-world problems. They also help us understand the behavior of functions and their relationships.

What makes a tricky integral difficult to solve?

A tricky integral can be difficult to solve because it may involve complex functions, multiple variables, or unconventional limits of integration. It may also require advanced techniques or creativity to find a solution.

How can I improve my skills in solving tricky integrals?

To improve your skills in solving tricky integrals, it is important to have a strong understanding of the basic principles and techniques of integration. Practicing regularly and seeking help from a tutor or mentor can also be beneficial.

What are some common strategies for solving tricky integrals?

Some common strategies for solving tricky integrals include using substitution, integration by parts, partial fractions, and trigonometric identities. It can also be helpful to break the integral into smaller parts and use known formulas or properties to simplify the problem.

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