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I can't find a way to create a working partial fraction to solve the following integral: $$\int \frac{n}{n^3+5}dn$$

This is part of the training to solve other kind of integrals where it's not so easy to find a partial fraction, so please bear with me if I try it on that and not on an easier integral.

Can someone show me how to solve that?

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marked as duplicate by Xander Henderson, Zacky, José Carlos Santos, KReiser, YuiTo Cheng yesterday

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    $\begingroup$ in general $a^3 + b^3 = (a+b) (a^2 - ab + b^2).$ here take symbols $a=n$ and $b = 5^{1/3}$ $\endgroup$ – Will Jagy Jul 6 at 21:05
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    $\begingroup$ Welcome to Maths SX! Is $n$ an integer? $\endgroup$ – Bernard Jul 6 at 21:05
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    $\begingroup$ I'm guessing that "defraction" refers to factoring. The denominator does not factor over the rational numbers, but that does not matter, it factors over the real numbers $\endgroup$ – Will Jagy Jul 6 at 21:08
  • $\begingroup$ I don't think there is anything else to it but to apply the factorization in Will Jagy's hint. The coefficients will be a bit hairy. Get on with it! $\endgroup$ – Jyrki Lahtonen Jul 6 at 21:08
  • $\begingroup$ And this question is related (and might be a better dupe target). $\endgroup$ – Xander Henderson yesterday
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$\textbf{Hint:}$ notice that $$x^3+5 = (x+5^{1/3})(x^2-5^{1/3}x+5^{2/3})$$ So, $$\frac{x}{x^3+5} = \frac{A}{x+5^{1/3}}+\frac{Bx+C}{x^2-5^{1/3}x+5^{2/3}}$$ for some constants $A$, $B$ and $C$.

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Here's a different approach, actually it doesn't use partial fractions, or maybe we can call it pseudo-partial fraction.

Start by letting $n=\sqrt[3]{5} x$ then: $$I=\int\frac{n}{n^3+5}dn=\frac{1}{\sqrt[3]{5}}\int\frac{x}{x^3+1}dx$$ Now we will substitute $x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt$.

The reason behind this substitution is that it gets rid of the third powers from the denominator, since: $(1+t)^3+(1-t)^3=6t^2+2$.

$$\Rightarrow I=-\frac{1}{\sqrt[3]{5}}\int \frac{1-t}{\left(\frac{1-t}{1+t}\right)^3+\underbrace{\frac{(1+t)^3}{(1+t)^3}}_{=1}}\frac{2}{(1+t)^3}=\frac{1}{\sqrt[3]{5}}\int \frac{t-1}{3t^2+1}dt$$ $$=\frac{1}{6\sqrt[3]{5}}\int\frac{6 t}{3t^2+1}dt-\frac{1}{\sqrt[3]{5}}\int\frac{1}{3t^2+1}dt$$ $$=\frac{1}{6\sqrt[3]{5}}\ln(3t^2+1)-\frac{1}{\sqrt[3]{5}\sqrt 3}\arctan(\sqrt 3 t)+C,\quad t=\frac{\sqrt[3]{5}-n}{\sqrt[3]{5}+n}$$

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  • $\begingroup$ Hi. thx you but plz give me an hint how you could solve that to get the same result again, because if i set for $x = \frac{n}{\sqrt[3]5}$ in the equation i get the result of $\int \frac{n \sqrt[3]{5}}{n^3+5}$. The rest is amazing and thx for that way to solve it $\endgroup$ – Eduard Tester Jul 7 at 8:15
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    $\begingroup$ @EduardTester You don't just replace $x$ with $\frac{n}{\sqrt[3]{5}}$ in order to get back to that, this is an integral, you must substitute $x=\frac{n}{\sqrt[3]{5}} \Rightarrow dx=\frac{dn}{\sqrt[3]{5}}$.$$\frac{1}{\sqrt[3]{5}}\int \frac{x}{x^3+1}dx=\frac{1}{\sqrt[3]{5}}\int \frac{\frac{n}{\sqrt[3]{5}}}{\frac{n^3}{5}+1}\frac{dn}{\sqrt[3]{5}}=\int \frac{n}{n^3+5}dn$$ $\endgroup$ – Zacky Jul 7 at 8:30
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    $\begingroup$ $\int \frac{n}{n^3+5} dn \Leftrightarrow \int \frac{n}{n^3+5} \sqrt[3]{5}dx = \int \frac{ \sqrt[3]{5}x} { (\sqrt[3]{5})^3 x^3+ (\sqrt[3]{5})^3} \sqrt[3]{5}dx = \int \frac{(\sqrt[3]{5})^2x}{(\sqrt[3]{5})^3(x^3+1)}dx = \int \frac{1}{\sqrt[3]{5}} \frac{x}{x^3+1}dx$ now i got it :) sry. i am a beginner atm and integrating is really hard if you dont have much experience $\endgroup$ – Eduard Tester Jul 7 at 17:55
  • $\begingroup$ can i ask you something? how did you get the idea to set $\frac{1-t}{1+t}$? is it just long life training to see the parts which xould be made much more easier? Can i ask another question? If you integrate 2 integrals like in that part does not creats 2 different constants to have +2C or C + D od did you create then just onb the secound integration the D part? so many questions :S $\endgroup$ – Eduard Tester Jul 8 at 12:36
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    $\begingroup$ all right. thx a lot. now i understand that a little bit better then before :) i have my problems with that tan, ctan, and co, but the day will come!!!!(invader zim of the integrals :D) if i also solve that kind of intergrals like a charm. and i like your style how you solve that integral. brutal for an newbi but also nice to see new ways :) br $\endgroup$ – Eduard Tester Jul 8 at 13:23

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