Can You Solve These Advanced Calculus Integrals?

In summary, according to the PF, this question is not allowed as a brain teaser because it is homework-like.
  • #1
Ratio Test =)
19
0
Hello :)
Here we go! :


1. [tex]
\int \frac{dx}{x^3+x^2+x+1}
[/tex]

2. [tex]
\int_{\frac{\pi}{2}}^{\pi} \left( sin(x) ln(x) - \frac{cos(x)}{x}\right) dx
[/tex]

Do your best :)
 
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  • #2
1 + x + x^2 + x^3 = (1-x^4)/(1-x)

So, we have to integrate -(x-1)/(x^4 - 1)

The singularities of this function are at x = -1, i and -i, so the partial fraction decomposition is:

-1/2 1/(x+1) +1/4 (1+i)/(x-i) + 1/4 (1-i)/(x+i)

The integral is thus given by:

-1/2 Log(x+1) + 1/4 (1+i)Log(x-i) + cc of last term. =

-1/2 Log(x+1) + 1/2 Re[(1+i)Log(x-i)]

Real and imaginary parts of logarithms of complex arguments are easily obtained as follows. We have:

Log[r exp(i theta)] = Log(r) + i theta

This means that:

Log(x + i y) = 1/2 Log(x^2 + y^2) + i arctan(y/x)

The integral is thus given by:

-1/2 Log(x+1) + 1/4 Log(x^2 + 1) + 1/2 arctan(1/x)

-1/2 Log(x+1) + 1/4 Log(x^2 + 1) - 1/2 arctan(x)

(pi absorbed in integration constant which we don't write down)
 
Last edited:
  • #3
Um, I believe #1 is just partial fractions, while #2 is integration by parts. No need to be fancy :)
 
  • #4
sin(x)ln(x)-cos(x)/x=(-cos(x))'ln(x)+(-cos(x))(ln(x))'=(-cos(x)ln(x))'
 
  • #5
Ratio Test =) said:
Hello :)
Here we go! :


1. [tex]
\int \frac{dx}{x^3+x^2+x+1}
[/tex]

2. [tex]
\int_{\frac{\pi}{2}}^{\pi} \left( sin(x) ln(x) - \frac{cos(x)}{x}\right) dx
[/tex]

Do your best :)

Welcome to the PF, Ratio. We generally do not allow homework-like brain teasers here on the PF, for obvious reasons. Please do not post this type of question again. I've moved this question to the Homework Help forums, and the normal homework rules apply (we don't do these types of questions for students).
 
  • #6
berkeman said:
Welcome to the PF, Ratio. We generally do not allow homework-like brain teasers here on the PF, for obvious reasons. Please do not post this type of question again. I've moved this question to the Homework Help forums, and the normal homework rules apply (we don't do these types of questions for students).


One Question : Who told you this is for my homework ?
Did you read the thread's title ?
 
  • #7
Ratio Test =) said:
One Question : Who told you this is for my homework ?
Did you read the thread's title ?

Don't imply I'm an idiot. If you don't understand why homework questions aren't allowed as brain teasers, please think about it a bit more. And re-read the PF Rules link at the top of the page that you agreed to when you joined here.

Thread locked. Show some brains people.
 

FAQ: Can You Solve These Advanced Calculus Integrals?

What are challenging integrals?

Challenging integrals are mathematical expressions that involve the calculation of the area under a curve or the volume of a three-dimensional shape. They often require advanced techniques and creative approaches to solve.

Why are challenging integrals important?

Challenging integrals are important because they are often encountered in real-world applications, such as physics, engineering, and economics. They also help to develop problem-solving skills and critical thinking abilities.

How do you approach solving challenging integrals?

There is no one set approach for solving challenging integrals. However, some common techniques include substitution, integration by parts, and partial fractions. It also helps to have a strong understanding of basic integration rules and concepts.

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Some common mistakes to avoid when solving challenging integrals include forgetting to account for constants, making algebraic errors, and not checking for extraneous solutions. It is also important to carefully check the final answer and see if it makes sense in the context of the problem.

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Practicing challenging integrals can help to improve problem-solving skills, develop critical thinking abilities, and deepen understanding of integration concepts. It can also prepare individuals for more advanced math courses and real-world applications.

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