January 29th, 2017, 09:15 PM  #1 
Newbie Joined: Jan 2017 From: Vancouver Posts: 1 Thanks: 0  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+(x9)/(x^2  2x+5). This gives me: ∫ x+2+(x9)/(x^2  2x+5) dx. Then I did the completing the square for the denominator (x^2  2x+5) and got (x1)^2 + 4. Now the integral becomes: ∫ x+2+(x9)/((x1)^2 + 4) dx. AT this stage I do not know how to proceed doing an integration of rational function.. If the denominator was like (x2)(x+1), then I would know how to proceed. I was able to do up to this point: (x^2)/2 + 2x +∫(x9)/((x1)^2 + 4) dx. 
January 29th, 2017, 09:27 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,382 Thanks: 1281 
these templates might help $\displaystyle{\int}~\dfrac{a}{\left(\dfrac{xb}{c}\right)^2+1}~dx = a c \tan ^{1}\left(\dfrac{bx}{c}\right)$ $\displaystyle{\int}~\dfrac{a x}{\left(\dfrac{xb}{c}\right)^2+1}~dx =a \left(\dfrac{1}{2} c^2 \log \left((bx)^2+c^2\right)b c \tan ^{1}\left(\frac{bx}{c}\right)\right)$ 
January 30th, 2017, 04:39 PM  #3 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
$\displaystyle \frac{x^{3}+1}{x^{2}2x+5}=x+2 \frac{x+9}{x^{2}2x+5}$ by long division. Use partial fractions to integrate last fraction. 
January 30th, 2017, 05:03 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,382 Thanks: 1281  Quote:
the two terms in the last fraction have to be split and massaged to match the templates I posted.  
January 30th, 2017, 05:31 PM  #5  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
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.  
January 30th, 2017, 06:11 PM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,622 Thanks: 2611 Math Focus: Mainly analysis and algebra 
$$\frac{x+9}{x^22x+5}$$ Differentiate the denominator: $\displaystyle \frac{\mathrm d}{\mathrm dx}\left( x^22x+5 \right) = 2x2$ Now spilt the numerator into multiples of the derivative and a constant remainder: $\displaystyle x + 9 = \tfrac12 (2x2) + 10$ So we have $$\frac{x+9}{x^22x+5} = \frac12 \cdot \frac{2x2}{x^22x+5} + \frac{10}{x^22x+5}$$ Since the denominator doesn't factorise, we complete the square on the second term: $$\frac{x+9}{x^22x+5} = \frac12 \cdot \frac{2x2}{x^22x+5} + \frac{10}{x^22x+1+4} = \frac12 \cdot \frac{2x2}{x^22x+5} + \frac{10}{(x1)^2+2^2}$$ Form the integral: $$\int \frac{x+9}{x^22x+5} \,\mathrm dx = \frac12 \int \frac{2x2}{x^22x+5} \,\mathrm dx + 10 \int \frac{1}{(x1)^2+2^2}$$ 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 $u=x^22x+5$ suffices. The second integral, under the substitution $2\tan v = x1$ resolves easily. If you haven't learned trigonometric substitutions yet, the substitution $v = \frac12(x1)$ reduces it to the integral $\displaystyle \int \frac{1}{v^2+1}\,\mathrm dv$ which you may know. 
January 30th, 2017, 06:38 PM  #7 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
The inuendos were unnecessary. I think we can assume Romsek got it.

January 30th, 2017, 06:43 PM  #8  
Senior Member Joined: Sep 2015 From: USA Posts: 2,382 Thanks: 1281  Quote:
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.  
January 30th, 2017, 06:52 PM  #9  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
 
January 30th, 2017, 07:00 PM  #10 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond  

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