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I'm trying to integrate this but I think the function can't be integrated? Just wanted to check, and see if anyone is able to find the answer (I used integration by parts but it doesn't work). Thanks in advance; the function I need to integrate is $$\int\frac{x}{x^5+2}dx$$

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  • $\begingroup$ Whoever asked you to integrate this has no mercy. The answer given by Wolfram is quite complicated ... $\endgroup$ – uniquesolution Oct 24 '15 at 21:06
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Probably you'd want to use the calculus of residues to do this.

But below I do it using first-year calculus methods.

The cumbersome part may be the algebra, and that's what I concentrate on here. \begin{align} x^5 + 2 & = \left( x+\sqrt[5]{2} \right) \underbrace{\left( x - \sqrt[5]{2} e^{i\pi/5}\right)\left( x - \sqrt[5]{2} e^{-i\pi/5}\right)}_\text{conjugates}\ \underbrace{\left( x - \sqrt[5]{2} e^{i3\pi/5}\right)\left( x - \sqrt[5]{2} e^{-i3\pi/5}\right)}_\text{conjugates} \\[10pt] & = \left( x+\sqrt[5]{2} \right) \left( x^2 - 2 \sqrt[5]{2} \cos\frac{\pi} 5 + \sqrt[5]{2}^2 \right)\left( x^2 - 2\sqrt[5]{2}\cos\frac{3\pi}5 + \sqrt[5]{2}^2 \right) \end{align} Then use partial fractions.

That the polynomials $x^2 - 2\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2$ and $x^2 - 2\sqrt[5]{2}\cos\frac{3\pi} 5 + \sqrt[5]{2}^2$ cannot be factored using real numbers can be seen from the way we factored them above. So you'll have $$ \cdots + \frac{Bx+C}{x^2 - 2\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2} + \cdots $$ etc. and you'll need to find $B$ and $C$.

Let $u=x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2$ then $du = (2x + \sqrt[5]{2}\cos\frac{\pi}5)\,dx$ and so $$ \frac{Bx + C}{x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2}\,dx = \underbrace{\frac{\frac B 2 \left( 2x+\sqrt[5]{2}\cos\frac{\pi}5 \right)}{x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2}\,dx}_\text{Use the substitution for this part.} + \underbrace{\frac{\left( 1 - \frac B2 \right) \sqrt[5]{2}\cos\frac{\pi} 5}{x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + 1}\,dx}_\text{See below for this part.} $$

To integrate $\frac{\left( 1 - \frac B2 \right) \sqrt[5]{2}\cos\frac{\pi} 5}{x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2}\,dx$, complete the square:

\begin{align} & x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2 = \left( x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2\cos^2\frac{\pi}5 \right) + \sqrt[5]{2}^2 \sin^2\frac{\pi} 5 \\[10pt] = {} & \left( x - \sqrt[5]{2}\cos\frac{\pi} 5 \right)^2 + \sqrt[5]{2}^2\sin^2 \frac{\pi} 5 \\[10pt] = {} & \left(\sqrt[5]{2}^2 \sin^2 \frac{\pi} 5 \right) \left( \left( \frac{x-\sqrt[5]{2}\cos\frac{\pi} 5}{\sqrt[5]{2}\sin\frac{\pi} 5} \right)^2 + 1 \right) = (\text{constant})\cdot(w^2 + 1) \\[10pt] & \text{and } dw = \frac{dx}{\sqrt[5]{2}\sin\frac{\pi}5}. \end{align} So you get an arctangent from this term.

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  • $\begingroup$ First line: every root should have modulus $\sqrt[5]{2}$, not $1$ as four of the "roots" in the RHS. $\endgroup$ – Did Oct 24 '15 at 21:17
  • $\begingroup$ It's easier to first rescale the variable $x$ t0 get rid of all the annoying $\sqrt[5]2$ factors. $\endgroup$ – Yves Daoust Oct 24 '15 at 21:24
  • $\begingroup$ @YvesDaoust Indeed! $\endgroup$ – Did Oct 26 '15 at 7:45
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You can do a partial fraction decomposition of $\frac{x}{x^5+2}$ which results in a very ugly result of the form

$$ \frac{x}{x^5+2} = \frac{a_1 x + a_2}{a_3x^2 + a_4x + a_5} + \frac{a_6 x + a_7}{a_8x^2 + a_9x + a_{10}} + \frac1{b_1 + xb_2}, $$ where all the $a$ an $b$ are some really ugly constants. Now you apply the linearity of the integral. The first two terms can be integrated using standard methods, eventually using $(\operatorname{atan}(x))' = \frac1{1+x^2}$. The last term should be abvious and will contain the natural logarithm at the end.

All in all, the answer will look very, as I stated already, ugly.

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