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January 29th, 2017, 09:15 PM   #1
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Integration of Rational Function Help

∫ (x^3 + 1)/(x^2 - 2x+5) dx.

First I did a long division, and the result was: x+2+(-x-9)/(x^2 - 2x+5).
This gives me: ∫ x+2+(-x-9)/(x^2 - 2x+5) dx.

Then I did the completing the square for the denominator (x^2 - 2x+5) and got (x-1)^2 + 4.
Now the integral becomes: ∫ x+2+(-x-9)/((x-1)^2 + 4) dx.

AT this stage I do not know how to proceed doing an integration of rational function.. If the denominator was like (x-2)(x+1), then I would know how to proceed. I was able to do up to this point:

(x^2)/2 + 2x +∫(-x-9)/((x-1)^2 + 4) dx.
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January 29th, 2017, 09:27 PM   #2
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these templates might help

$\displaystyle{\int}~\dfrac{a}{\left(\dfrac{x-b}{c}\right)^2+1}~dx = -a c \tan ^{-1}\left(\dfrac{b-x}{c}\right)$

$\displaystyle{\int}~\dfrac{a x}{\left(\dfrac{x-b}{c}\right)^2+1}~dx =a \left(\dfrac{1}{2} c^2 \log \left((b-x)^2+c^2\right)-b c \tan ^{-1}\left(\frac{b-x}{c}\right)\right)$
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January 30th, 2017, 04:39 PM   #3
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$\displaystyle \frac{x^{3}+1}{x^{2}-2x+5}=x+2 -\frac{x+9}{x^{2}-2x+5}$

by long division. Use partial fractions to integrate last fraction.
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January 30th, 2017, 05:03 PM   #4
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Quote:
Originally Posted by zylo View Post
$\displaystyle \frac{x^{3}+1}{x^{2}-2x+5}=x+2 -\frac{x+9}{x^{2}-2x+5}$

by long division. Use partial fractions to integrate last fraction.
really? Did you even try?



the two terms in the last fraction have to be split and massaged to match the templates I posted.
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January 30th, 2017, 05:31 PM   #5
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Quote:
Originally Posted by romsek View Post
really? Did you even try?
the two terms in the last fraction have to be split and massaged to match the templates I posted.
What? Why bother and learn nothing but maybe get the answer- 0 points.

If you look up partial fractions you try to factor denominator into 2 linear factors and discover you can't. So you just write
$\displaystyle \frac{x+9}{x^{2}-2x+5}=\frac{x}{x^{2}-2x+5}+\frac{9}{x^{2}-2x+5}$
which you can look up in a table of integrals.

You have thus learned:
1) If degree of numerator is greater than denominator, divide to get a fraction with smaller degree in the numerator and use partial fractions.
2) If you can't factor beyond a quadratic in the denominator, you get terms like above which are standard integrals.
10 points.
Explanation with right answer- 10 points

Last edited by zylo; January 30th, 2017 at 05:34 PM.
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January 30th, 2017, 06:11 PM   #6
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$$\frac{x+9}{x^2-2x+5}$$

Differentiate the denominator: $\displaystyle \frac{\mathrm d}{\mathrm dx}\left( x^2-2x+5 \right) = 2x-2$

Now spilt the numerator into multiples of the derivative and a constant remainder: $\displaystyle x + 9 = \tfrac12 (2x-2) + 10$

So we have $$\frac{x+9}{x^2-2x+5} = \frac12 \cdot \frac{2x-2}{x^2-2x+5} + \frac{10}{x^2-2x+5}$$

Since the denominator doesn't factorise, we complete the square on the second term: $$\frac{x+9}{x^2-2x+5} = \frac12 \cdot \frac{2x-2}{x^2-2x+5} + \frac{10}{x^2-2x+1+4} = \frac12 \cdot \frac{2x-2}{x^2-2x+5} + \frac{10}{(x-1)^2+2^2}$$

Form the integral:
$$\int \frac{x+9}{x^2-2x+5} \,\mathrm dx = \frac12 \int \frac{2x-2}{x^2-2x+5} \,\mathrm dx + 10 \int \frac{1}{(x-1)^2+2^2}$$

The first integral is of the form $\displaystyle \int \frac{f'(x)}{f(x)} \,\mathrm dx$, which is a form you should always look out for. If you haven't learned it yet, the substitution $u=x^2-2x+5$ suffices.

The second integral, under the substitution $2\tan v = x-1$ resolves easily. If you haven't learned trigonometric substitutions yet, the substitution $v = \frac12(x-1)$ reduces it to the integral $\displaystyle \int \frac{1}{v^2+1}\,\mathrm dv$ which you may know.
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January 30th, 2017, 06:38 PM   #7
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The inuendos were unnecessary. I think we can assume Romsek got it.
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January 30th, 2017, 06:43 PM   #8
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Quote:
Originally Posted by zylo View Post
What? Why bother and learn nothing but maybe get the answer- 0 points.

If you look up partial fractions you try to factor denominator into 2 linear factors and discover you can't. So you just write
$\displaystyle \frac{x+9}{x^{2}-2x+5}=\frac{x}{x^{2}-2x+5}+\frac{9}{x^{2}-2x+5}$
which you can look up in a table of integrals.

You have thus learned:
1) If degree of numerator is greater than denominator, divide to get a fraction with smaller degree in the numerator and use partial fractions.
2) If you can't factor beyond a quadratic in the denominator, you get terms like above which are standard integrals.
10 points.
Explanation with right answer- 10 points
maybe in your classes.

In the classes I took, the whole point of an exercise like this was to mangle the original expression into a form that could easily be integrated from the table of simple integral forms.

We weren't expected, in problems such as this one, to derive the forms.

In some other problems we were expected to go ahead and derive some specific integral form but it was made clear that that was what the problem was about.

You don't have to reinvent the wheel for every problem.
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January 30th, 2017, 06:52 PM   #9
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Quote:
Originally Posted by romsek View Post
maybe in your classes.

In the classes I took, the whole point of an exercise like this was to mangle the original expression into a form that could easily be integrated from the table of simple integral forms.
Thank you, that's exactly what I did, though I would hardly call it "mangling." Sorry a little elementary algebra is causing you and your friend so much trouble.

Quote:
Originally Posted by zylo View Post
$\displaystyle \frac{x^{3}+1}{x^{2}-2x+5}=x+2 -\frac{x+9}{x^{2}-2x+5}$

by long division. Use partial fractions to integrate last fraction.
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January 30th, 2017, 07:00 PM   #10
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Quote:
Originally Posted by zylo View Post
The inuendos were unnecessary. I think we can assume Romsek got it.
Threads on this forum are not for you to appropriate, as you are in the habit of doing. I'm imposing a 7-day ban.
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