Помогите взять интеграл

[ins-]

Помогите взять интеграл:
\int \frac{3x^{4}+12x^{3}+6}{x^{5}+5x^{4}+16x+12} \,dx

P.S. The source of this integral is sosmath.com it is also not solved in mathlinks.ro. I believe it is interesting integral. I had solved 3 similar integrals - they sometimes requires special approaches - substitution+interesting simplifying or not defined coefficients, but I have no idea for this concrete integral.

[Андрей123]

Попробуйте разложить дробь на простейшие(один из корней знаменателя равен -1). Это, конечно, не очень приятно, но зато надёжно.

[незваный гость]

I am sure what kind of answer do you expect. The approach proposed by Андрей123 is the right one. The answer — $\frac{3}{5}\ln(q(x)) - \frac{18}{5} \sum\limits_{\xi: q(\xi)=0} \frac{ln(x-\xi)}{q'(\xi)} +C$, where $q(x) = x^{5}+5x^{4}+16x+12$ can be converted in several equivalent forms (using propeties of roots).

E.g., $-3\ln(x+1)+\frac{3}{3514}\sum\limits_{\xi:q(\xi)/(\xi+1)=0} (195\xi^3+576\xi^2-1371\xi+2975) \ln(x-\xi) + C$,

Do you expect “a nice” expression? If so, why do you?

[ins-]

I think a "nice" expression may be found. I don't understand your solution in depth, but I think - we may try to find the answer as
ln(P(x)/Q(x)) where P(x), Q(x) are some polynomials of x using the method of not defined coefficients. May we use this idea? What is the "exact" function by your way "незванны гость"?

[незваный гость]

незваный гость писал(а):

Do you expect “a nice” expression? If so, why do you?

ins- писал(а):

What is the "exact" function by your way

What do you mean by that?
$\frac{3}{5}\ln(q(x)) - \frac{18}{5} \sum\limits_{\xi: q(\xi)=0} \frac{ln(x-\xi)}{q'(\xi)} +C$ is a full-defined function.

[ins-]

I don't understand your answer ... don't expect lots from me - math is only my hobby. It is what I don't understand.
What do you mean by that?
$\frac{3}{5}\ln(q(x)) - \frac{18}{5} \sum\limits_{\xi: q(\xi)=0} \frac{ln(x-\xi)}{q'(\xi)} +C$ is a full-defined function.[/quote]
I believe it may be found a good expression because for the integral:
$ \int \frac{x^{4} + 4x^{2}-1}{2x^{6}-3x^{4}+4x^{2}+1} \,dx $
and two similar "ugly" integrals there exist such expressions. My problem is difficult but I believe a "good" expression exist.

[незваный гость]

незваный гость писал(а):

$\sum\limits_{\xi: q(\xi)=0}$

The sum is by all roots of the polynomial $q(x)=0$.

ins- писал(а):

I believe it may be found a good expression because …

… Also, because 1, 2, 3, 4, 5, 6 divide 60, I believe that any integer divides 60. And I habitually call problems I cannot solve difficult.

Sorry for an explosion of sarcasm.

[ins-]

Yes, maybe you are right. How you get your answer? By using software for integration or by some other way? If it is common way where I can get more information? May your approach be used for my second integral? What is the answer of my second integral on your opinion?

P.S. You may be right, but just to know, the equation: "denominator=0" have two complex roots.

[незваный гость]

ins- писал(а):

$\int \frac{3x^{4}+12x^{3}+6}{x^{5}+5x^{4}+16x+12} \,dx $

We can represent $\frac{p(x)}{q(x)}$ as a sum of simple fractions (if degree of $ p ( x ) $ is less than degree of $q(x)$): $\frac{p(x)}{q(x)} = \sum_\xi \frac{p(\xi)}{q'(\xi)(x-\xi)}$, where sum is by the roots of $q(x)$. In our case it happened that $p(x) = \frac{3}{5}q'(x)-\frac{18}{5}$. Integrating, we have $\int = $ $\frac{3}{5} \sum_\xi \ln(x-\xi)-\frac{18}{5} \sum_\xi \frac{\ln(x-\xi)}{q'(\xi)}+C$. We can simplify $\sum_\xi \ln(x-\xi)= $ $\ln(\prod_\xi (x-\xi))=$ $\ln(q(x))$. So, finally $\int = \frac{3}{5} \ln(q(x))-\frac{18}{5}\sum_\xi \frac{\ln(x-\xi)}{q'(\xi)}+C$.

ins- писал(а):

$ \int \frac{x^{4} + 4x^{2}-1}{2x^{6}-3x^{4}+4x^{2}+1} \,dx $

Here we are dealing with a different opportunity: ${2x^{6}-3x^{4}+4x^{2}+1}$ can be factored as ${2x^{6}-3x^{4}+4x^{2}+1} = q_1(x) q_2(x)$. What happens next is that $\frac{x^{4} + 4x^{2}-1}{2x^{6}-3x^{4}+4x^{2}+1} = $ $\frac{p_1(x)}{q_1(x)} + \frac{p_2(x)}{q_2(x)}$, and (surprise! surprise!) $p_1(x) = a_1 q_1'(x)$, while $p_2(x) = a_2 q_2'(x)$. Therefore, $\int = a_1 \ln(q_1(x)) + a_2 \ln(q_2(x))$.

Добавлено спустя 3 минуты 19 секунд:

ins- писал(а):

You may be right, but just to know, the equation: "denominator=0" have two complex roots.

What are you talking about?!

[ins-]

Please, take a look at this solution:

$\left(arctg\left(\frac {P}{Q}\right)\right)^{'} = \frac {1}{1 + (\frac {P}{Q})^{2}}.\left(\frac {P}{Q}\right)^{'} = \frac {1}{1 + (\frac {P}{Q})^{2}}.\frac {P^{'}.Q - PQ^{'}}{Q^{2}} = \frac {P^{'}Q - P.Q^{'}}{P^{2} + Q^{2}}$

We'll try to find P, Q, such that: $ (*) P^{2} + Q^{2} = 2x^{6} - 3x^{4} + 4x^{2} + 1 $.
We'll try to find P in the form: $ \sqrt {2}(x^{3} + ax); Q = bx^{2} + c $ where $ a, b, c $ are not defined coefficients.
From (*) follows that the following equations are right: $ 4a + b^{2} = - 3, 2a + 2bc = 4, c^{2} = 1 $.
After solving them as a system we choose one good solution: $ a = - 1, b = 1, c = 1 $.
So our functions are: $ P = \sqrt {2}(x^{3} - x), Q = x^{2} + 1 $.
$ P^{'}Q - P.Q^{'} = \sqrt {2}(x^4 + 4x^2 + 1) $.
It means that $ arctg\left(\frac {P}{Q}\right)^{'} = \frac {1}{\sqrt {2}}.\frac {P^{'}Q - P.Q^{'}}{P^{2} + Q^{2}} = \frac {x^{4} + 4x^{2} - 1}{2x^{6} - 3x^{4} + 4x^{2} + 1} $
So the answer is: $ \int \frac {x^{4} + 4x^{2} - 1}{2x^{6} - 3x^{4} + 4x^{2} + 1}\,dx = \frac {1}{\sqrt {2}}arctg\left(\frac {\sqrt {2}(x^{3} - x)}{x^{2} + 1}\right) + C $

My goal here is to understand is there some similar "focus" but not kidding you. Solving such integrals is something similar as try to solve some specific case of 5-th or 6-th degree equation with 'not random selected coefficients'. But unfortunately I cannot find such a 'focus'. It is an random but good looking integral for me from sosmath.com. I have two more similar integrals solved with some 'exotic' substitution. It was the reason for me to try solve this integral. Interesting moment with such integrals is that the denominator cannot be factored in some good looking way and general approach like described here sometimes doesn't work very well. Such kind of integrals are also a joke with integration software and may be used to show how difficult is to write a good integration software. Do you want to take a look of my other integrals?

I wanted to say that $ x^5 + 5x^4 + 16x + 12 = 0 $ have three reals and two complex roots.

[незваный гость]

ins- писал(а):

Please, take a look at this solution:
…
My goal here is to understand is there some similar "focus"

A trick is a trick exactly for the reason that it is not universal. One can expect that unsing representation of simple fractions he can always solve integral, but he cannot expect that there is a trick giving a nice-looking representation. I think to some degree it might be a fault of education system: students learn tricks without necessary stress that life is tough.

ins- писал(а):

Interesting moment with such integrals is that the denominator cannot be factored in some good looking way and general approach like described here sometimes doesn't work very well.

Actually, you are wrong on both counts. Firstly, the denominator is a polynomial of squares ($q(x) = q_*(x^2)$). It means, that with every root $\xi$ $(-\xi)$ is a root also. So we can group roots and factor $q(x) = (x-\xi_1)(x-\xi_2)(x-\xi_3)$ $ (x+\xi_1)(x+\xi_2)(x+\xi_3) = $ $(x^2+a x^2+bx +c)$ $(x^2-a x^2+bx -c)$. Although coefficients are complex, factors are quite nice-looking.

Secondly, I have shown above how general approach works, and why one can simplify result of general approach is this particular case.

ins- писал(а):

Such kind of integrals are also a joke with integration software and may be used to show how difficult is to write a good integration software.

I doubt that. They probably used to be, but not any more…

Note on the margin: $\ln(x+{\rm i}y) = \frac12\ln(x^2+y^2) + {\rm i}\arctg(\frac{y}{x})$.

[ins-]

Thank you for the time spent with this problem and valuable remarks.