I just encountered the following integral $$\int \frac{x^2}{1+x^5} \, dx $$ At first it appeared to be simple, but I don't know how to solve it. Please share any ideas.
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$\begingroup$ Quite a long closed form.... $\endgroup$– Prasun BiswasJun 30, 2015 at 1:41
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3$\begingroup$ Hint: partial fractions $\endgroup$– Ben GrossmannJun 30, 2015 at 1:49
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1$\begingroup$ @Omnomnomnom This is not looking like partial fraction problem , if you know please tell me how is it solvable through partial fraction. $\endgroup$– satyatechJun 30, 2015 at 1:55
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$\begingroup$ You could just integrate the terms one at a time $\endgroup$– Ben GrossmannJun 30, 2015 at 1:58
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$\begingroup$ Hint: Use partial fractions to change it into 3 fractions. The quadratic ones can be done either by partial fractions over $\Bbb{C}$, or by using a trig substitution. The other one can be done iwth the power rule. $\endgroup$– TeocJun 30, 2015 at 2:00
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1 Answer
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In this problem, the hard part is the algebra.
\begin{align} x^5 + 1 & = (x+1)(x^4-x^3+x^2-x+1) \\[10pt] & = (x+1)\left(x^2 - 2x\cos\frac\pi5 + 1\right)\left(x^2 - 2x\cos\frac{3\pi}5 + 1\right) \end{align}
These two quadratic factors are irreducible, as can be seen by the fact that their discriminants are negative.
Next, proceed to partial fractions. Completing the square, you get \begin{align} x^2 -2x\cos\frac\pi 5 +1 & =\left( x^2 - 2x\cos\frac\pi 5 + \cos^2\frac\pi5\right) + \sin^2 \frac\pi5 \\[10pt] & = \left( x - \cos\frac\pi5 \right)^2 +\sin^2\frac\pi 5. \end{align}
If you have $\dfrac{Ax+B}{\left( x - \cos\frac\pi5 \right)^2 +\sin^2\frac\pi 5} \, dx$, you can write it as \begin{align} & \frac{A\left(x-\cos\frac\pi5\right)}{\left( x - \cos\frac\pi5 \right)^2 +\sin^2\frac\pi 5}\,dx + \frac{B + A\cos\frac\pi 5}{\left( x - \cos\frac\pi5 \right)^2 +\sin^2\frac\pi 5}\,dx \\[10pt] = {} & \frac{\frac 1 2\,du}{u} + \frac{B + A\cos\frac\pi 5}{\left( x - \cos\frac\pi5 \right)^2 +\sin^2\frac\pi 5}\,dx \end{align} The first term yields a logarithm and the second an arctangent.
Moral: In this probelm, the hard part is the algebra.
So how did I get $\pi/5$ and $3\pi/5$?
The point is that $x^5+1=0$ iff $x^5 = -1$, and the $5$th roots of $-1$ are $\cos\dfrac\pi5 + i\sin\dfrac\pi 5$ and other points on the circle differing from that by a fifth of a circle, i.e. $2\pi/5$ radians. One of those points is $-1$, and that's where $(x+1)$ came from. Two of those points are $\cos\frac\pi5 \pm i\sin\frac\pi5$, and two are $\cos\frac {3\pi}5\pm i\sin\frac{3\pi}5$. So $$ \left(x - \cos\frac\pi 5 - i \sin\frac\pi5\right)\left(x - \cos\frac\pi 5 + i \sin\frac\pi5\right) = \left(x^2 - 2x\cos\frac\pi5+1\right). $$
