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How to compute this integral? I stuck at a point where I get $\displaystyle\int\frac{1}{t^5-1}+ \cdots $ $$\int\sqrt[5]{\frac{x+5}{x-5}}\,\mathrm dx$$ using $\displaystyle t=\sqrt[5]{\frac{x+5}{x-5}}$

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  • $\begingroup$ The evaluation of Dr. Sonnhard Graubner's rational integral from the answer below can be found here. $\endgroup$
    – Lucian
    Commented Dec 8, 2014 at 20:10

4 Answers 4

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Using $t = \sqrt[5]{\frac{x+5}{x-5}}$, you get $$ t^5 (x-5) = x+5 $$ $$ x(t^5 - 1) = 5 + 5t^5 $$ $$ x = \frac{5(t^5 + 1)}{t^5-1} $$ and thus $$ dx = 5\frac{(t^5-1)(5t^4)-(t^5+1)(5t^4)}{(t^5-1)^2}dt $$ $$ dx = \frac{-50t^4}{(t-1)^2(t^4 + t^3 + t^2 + t + 1)^2}dt $$ Thus the integral becomes $$ \int \frac{-50t^5}{(t-1)^2(t^4 + t^3 + t^2 + t + 1)^2}dt $$

The integrand is a rational function, and so we may integrate using Partial Fractions.

http://en.wikipedia.org/wiki/Partial_fraction_decomposition#Over_the_reals

Before we can do that, however, we have to factor the denominator fully. That is, we need to factor $(t^4 + t^3 + t^2 + t + 1)$.

If we plot the graph of $g(t) = t^4 + t^3 + t^2 + t + 1$, we see that $g(t)$ appears not to have any roots (zeros). We therefore conclude that $g(t)$ has no linear factors, in which case it factors over the reals as a product of two quadratic factors.

In fact, since the leading coefficient and constant coefficient of $g$ are both $1$, we might guess that $g(t) = (t^2 + Vt + 1)(t^2 + Wt + 1)$ for some values of $V$ and $W$. We set $g(t)$ equal to this factored expression and solve for $V$ and $W$ by equating coefficients, and we find that this guess is correct: $$ t^4 + t^3 + t^2 + t + 1 = t^4 + t^3(V + W) + t^2(1 + VW + 1) + t(V + W) + 1 $$ yields $$ V+W = 1 $$ $$ VW + 2 = 1 $$ Substituting the first equation $W = 1-V$ into the second yields $$ V - V^2 + 1 = 0 $$ This equation $V^2 - V - 1= 0$ has two solutions: $$ V = \frac{1\pm \sqrt{5}}{2} $$ Substituting either value in for $V$, we see that $W$ equals the other one: $$ W = 1 - V = 1 - \frac{1 \pm \sqrt{5}}{2} = \frac{2 - (1 \pm \sqrt{5})}{2} = \frac{1 \mp \sqrt{5}}{2} $$ Thus we get $$ g(t) = \big(t^2 + \frac{1+\sqrt{5}}{2}t + 1\big)\big(t^2+\frac{1-\sqrt{5}}{2}t + 1\big), $$ and we may easily check this result by multiplying out the two factors on the right hand side.

It turns out that the discriminant $(b^2 - 4ac)$ of both quadratic factors is negative, so as expected, neither one factors further into linear factors (over the reals). Notice that $\frac{1+\sqrt{5}}{2} = 1.618\ldots$ is the golden ratio, and is commonly referred to as $\phi$. Let's also write $\bar{\phi} = 1 - \phi = \frac{1-\sqrt{5}}{2}$.

Now we may use partial fractions:

$$ \frac{-50t^5}{(t-1)^2(t^2 +\phi t + 1)^2 (t^2 + \bar{\phi}t +1)^2} $$ $$ = \frac{A}{t-1} + \frac{B}{(t-1)^2} + \frac{Ct+D}{t^2 + \phi t + 1} + \frac{Et + F}{(t^2 + \phi t + 1)^2} + \frac{Gt + H}{t^2 + \bar{\phi}t + 1} + \frac{It + J}{(t^2 + \bar{\phi}t + 1)^2} $$

Clearing denominators, we get an equation of polynomials in $t$, and we may solve for the ten variables $A, B, \ldots, J$, by once again setting the corresponding coefficients equal to each other. We get a system of ten linear equations, corresponding to the coefficients of $t^0, t^1, \ldots t^9$, in ten variables. The left hand side of all of the equations will be zero, except for the equation corresponding to $t^5$, where the left hand side is -50.

Needless to say, it is a pain to solve this system by hand. One relation that might help you with this is that $\phi \bar{\phi} = -1$.

I used a computer to solve this system of equations. It says that $A=B=-2, C = 1+\sqrt{5}, D = 5-\sqrt{5}, E = -2\sqrt{5}, F = -5+\sqrt{5}, G = 1-\sqrt{5}$, $H = 5+\sqrt{5}$, $I = 2\sqrt{5}$, and $J = -5-\sqrt{5}$.

We can integrate $\int \frac{A}{t-1} dt = A\ln\lvert t-1 \rvert$ and $\int \frac{B}{(t-1)^2} dt = \frac{-B}{t-1}$ directly, but before integrating the other terms, we would complete the square in the denominator: $$ t^2 + \phi t + 1 = \big(t+\frac{\phi}{2}\big)^2 -\frac{\phi^2}{4} + 1 = \big(t + \frac{\phi}{2}\big)^2 + \big(\frac{4-\phi^2}{4}\big) $$ So we'll make two more substitutions: $$ u = t+\frac{\phi}{2}, $$ $$ v = t+ \frac{\bar{\phi}}{2}, $$ $$ du = dv = dt. $$ We need to integrate $$ \int \Big(\frac{C(u-\frac{\phi}{2}) + D}{u^2 + \frac{4-\phi^2}{4}} + \frac{E(u-\frac{\phi}{2}) + F}{(u^2 + \frac{4-\phi^2}{4})^2}\Big)du, $$ along with a very similar integral in $v$ (not shown). To solve this integral in $u$, let's set $p = \sqrt{\frac{4-\phi^2}{4}}$, which is real and positive since $4 > \phi^2$. Rearranging, this integral is now: $$ \int \Big(\frac{Cu}{u^2 + p^2} + \frac{D-\frac{C\phi}{2}}{u^2 + p^2} + \frac{Eu}{(u^2 + p^2)^2} + \frac{F-\frac{E\phi}{2}}{(u^2 + p^2)^2} \Big)du $$ These four fractions may now be integrated one at a time. The first and third are done by changing variables one more time, with $w = u^2+\frac{4-\phi^2}{4}$, so $\frac{dw}{2} = u\phantom{.} du$. The second and fourth may be done by the trig substitution $u = p \tan(\theta)$, with $du = p \sec^2(\theta)d\theta$.

http://en.wikipedia.org/wiki/Trigonometric_substitution#Integrals_containing_a2_.2B_x2

After integrating with respect to $\theta$ and $w$, substitute back using $\theta = \tan^{-1}\left(\frac{u}{ p}\right)$ and $w = u^2 + \frac{4-\phi^2}{4}$, then $u = t + \frac{\phi}{2}$, and finally $t = \sqrt[5]{\frac{x+5}{x-5}}$ to get the answer in terms of $x$.

Throw in a constant of integration, and you're done.

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Substituting $t=\sqrt[\Large5]{\frac{x+5}{x-5}}$, $$ \begin{align} \int\sqrt[\Large5]{\frac{x+5}{x-5}}\,\mathrm{d}x &=\int t\,\mathrm{d}\frac{10}{t^5-1}\\ &=\frac{10t}{t^5-1}-10\int\frac{\mathrm{d}t}{t^5-1}\\ \end{align} $$ Substituting $\sin(2\pi/5)u=t-\cos(2\pi/5)$ and $\sin(4\pi/5)v=t-\cos(4\pi/5)$, $$ \begin{align} &\frac1{t^5-1}\\ &=\frac15\sum_{k=0}^4\frac{e^{2\pi ik/5}}{t-e^{2\pi ik/5}}\\ &=\frac{2t\cos(2\pi/5)-2}{t^2-2t\cos(2\pi/5)+1}+\frac{2t\cos(4\pi/5)-2}{t^2-2t\cos(4\pi/5)+1}+\frac1{t-1}\\ &=2\frac{\cos(2\pi/5)(t-\cos(2\pi/5))-\sin^2(2\pi/5)}{(t-\cos(2\pi/5))^2+\sin^2(2\pi/5)}\\ &+2\frac{\cos(4\pi/5)(t-\cos(4\pi/5))-\sin^2(4\pi/5)}{(t-\cos(4\pi/5))^2+\sin^2(4\pi/5)}\\ &+\frac1{t-1}\\[6pt] &=2\frac{\cot(2\pi/5)u-1}{u^2+1}+2\frac{\cot(4\pi/5)v-1}{v^2+1}+\frac1{t-1} \end{align} $$ so that $$ \begin{align} \int\frac1{t^5-1}\,\mathrm{d}t &=\cos(2\pi/5)\log(u^2+1)-2\sin(2\pi/5)\arctan(u)\\ &+\cos(4\pi/5)\log(v^2+1)-2\sin(4\pi/5)\arctan(v)\\[6pt] &+\log(t-1)\\[6pt] &+C \end{align} $$ Therefore, using the substitutions above, which are unwieldy to write, but simple to compute, $$ \begin{align} \int\sqrt[\Large5]{\frac{x+5}{x-5}}\,\mathrm{d}x &=\frac{10t}{t^5-1}\\[3pt] &-10\cos(2\pi/5)\log(u^2+1)+20\sin(2\pi/5)\arctan(u)\\[6pt] &-10\cos(4\pi/5)\log(v^2+1)+20\sin(4\pi/5)\arctan(v)\\[6pt] &-10\log(t-1)\\[6pt] &-10\,C \end{align} $$

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setting $$t=\sqrt[5]\frac{x+5}{x-5}$$ we get $$x=\frac{5(t^5+1)}{t^5-1}$$ and we get $$dx=-\frac{50t^4}{(t-1)^2(t^4+t^3+t^2+t+1)^2}dt$$ thus our integral is $$-50\int\frac{t^5}{(t-1)^2(t^4+t^3+t^2+t+1)^2}dt$$ this is not so easy to solve but rational and can be solved explicitely

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    $\begingroup$ It's a little easier to see that final expression inside the integral as $\frac{1}{t^5-1} + \frac{1}{(t^5-1)^2}$ $\endgroup$ Commented Dec 8, 2014 at 18:53
  • $\begingroup$ Could you explain how the last one could be solved ? $\endgroup$
    – GorillaApe
    Commented Dec 8, 2014 at 19:03
  • $\begingroup$ @Parhs: I think that may be using imaginary fifth roots of unity will simplify the Partial Fraction Decomposition. $\endgroup$
    – user170039
    Commented Dec 16, 2014 at 12:45
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You are looking at Gauss Hypergeometric ${}_2F_1$. See here or here.

Compare e.g., $$\int \sqrt[5]{\frac{x+5}{x−5}}\,dx = \int{(x+1)^{\frac{1}{5}}(x-8+1)^{-\frac{1}{5}}}dx$$ with http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/07/01/01/.

I would not expect to find a "simpler" answer than one expressed in terms of ${}_2F_1$.

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