At first I had no doubt that I will have to use partial fractions on this integral: \begin{equation} \int \frac{3x^2+4}{x^5+x^3}dx \end{equation} I split it into two integrals and one of them give me back this equation: \begin{equation} \ A(x^2+1)+Bx(x^2+1)+Cx^2(x^2+1)+Dx^4+Ex^3 =1 \end{equation} Then if I check what happens if $x = 0$ I find out that $A = 1$ after that I checked what happens then $ x =1 \ \ x=-1 \ \ x=2 $ my results there : \begin{equation} \ 2B+E=0 \\ \ 2C+D=-1 \\ \ E+2D=2 \end{equation} I am realizing that I can joggle variables in any fashion I like I will not be get their values. So using partial fractions is not effective for this particular case. What other way could you suggest of handling this equation?

P.S. I may have made a calculation mistake. In that case I am sorry for wasting your time but I would appreciate if you could point out my mistake.

EDIT: I wrote 'I split it into two integrals' it seams it needs to be shown: \begin{equation} 3\int \frac{dx}{x^2+1}dx +4 \int \frac{dx}{x^5+x^3}dx = \\ \ \arctan x +4(\int \frac{A}{x^3}dx+\int \frac{B}{x^2}dx+\int \frac{C}{x}dx+\int \frac{Dx+E}{x^2+1}dx) \end{equation}

  • $\begingroup$ Why do you have $=1$ in the equation and not $=3x^2+4$? $\endgroup$ – Dennis Gulko Mar 1 '12 at 23:32
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    $\begingroup$ Once again: "solved" is the wrong word. One solves problems; one solves equations. One evaluates expressions. $\endgroup$ – Michael Hardy Mar 2 '12 at 0:20

I don't think it can be evaluated easily without partial fractions (see Mike's answer, though). After your edit, I see you set things up correctly with the integral you're using partial fractions on. But you only have three equations (which I haven't checked). You have four unknowns. Find another equation.

As for the first integral, though, it should be $3\int {dx\over x^3+x}$...

Let's try a decomposition at the start without splitting the integral up first (it would be just as easy): $$ {3x^2+4\over x^3(x^2+1)}= {A\over x}+{B\over x^2}+{C\over x^3}+{Dx+E\over x^2+1} $$ Which gives $$ 3x^2+4=Ax^2(x^2+1)+Bx(x^2+1)+C(x^2+1)+(Dx+E)x^3. $$ Let's figure out what we can be giving $x$ particular values: Setting $x=0$ gives $C=4$. No more nice values. But we can write now

$$ 3x^2+4=Ax^2(x^2+1)+Bx(x^2+1)+4(x^2+1)+(Dx+E)x^3. $$ Expanding the right hand side of the above and putting it in standard form: $$ 3x^2 +4=(A+D)x^4+(E+B)x^3+(A+4)x^2+Bx+4 $$ So, equating coefficients: $$\eqalign{ A&=-D\cr E&=-B\cr A&=-1\cr B&=0;\cr } $$ whence $A=-1$, $D=1$, $E=B=0$, and (from before) $C=4$.

So, $$ {3x^2+4\over x^3(x^2+1)}= {-1\over x}+ {4\over x^3}+{ x \over x^2+1}. $$

  • $\begingroup$ Ok I promise I will not ask any more question till I get some sleep. Thanks for your patience. $\endgroup$ – Povylas Mar 1 '12 at 23:52

To answer the question as asked, it appears given the factor of $1+x^2$ in the denominator, if you want to avoid partial fractions, the substituion

$x=\tan u,dx=\sec^2udu$

looks like it would simplify the denominator.




$\int3\cot u+4\cot^3udu=\int3\cot u+4\cot u(\csc^2u-1)du=$

$\int-\cot u+4\cot u\csc^2udu=$

$-\ln|\sin u|-2\cot^2u+C$

That first resubstitution is going to be ugly.


So it looks like we have our result as $-\ln\sqrt{\frac{x^2}{x^2+1}}-\frac2{x^2}+C$

  • $\begingroup$ Nice. A typo: in the fifth equation, you should have $\cot^2 u=\csc^2 u-1$. $\endgroup$ – David Mitra Mar 2 '12 at 1:43
  • $\begingroup$ Oops. I'll have to fix that. $\endgroup$ – Mike Mar 2 '12 at 2:13
  • $\begingroup$ @David Does it look all right now? I mean, I could possibly rewrite the ln term, but are there any other errors I missed? $\endgroup$ – Mike Mar 2 '12 at 2:23
  • $\begingroup$ It looks correct to me now. But, I have been known to make errors myself :) $\endgroup$ – David Mitra Mar 2 '12 at 2:28

The demoninator factors into $x^3(x^2+1)$, so to use partial fractions you will end up with $\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{Dx+E}{x^2+1}$.


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