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In this post, I shall focus on how I thought instead of the detailed solutions which I had posted individually for those who are interested.

Let’s start with the easier first.

$(1)\quad$ When $k=3$, $\displaystyle \quad I _ { 3 } = \int \frac { x ^ { 3 } } { 1+ x ^ { 8 } } d x = \frac { 1 } { 4 } \int \frac { d ( x ^ { 4 } ) } { 1 + ( x ^ { 4 } ) ^ { 2 } } = \frac { 1 } { 4 } \tan ^ { - 1 } ( x ^ { 4 } ) + C $

$(2)\quad $When $k=7$, $\displaystyle \quad I _ { 7 } = \int \frac { x ^ { 7 } } { 1 + x ^ { 8 } } d x = \frac { 1 } { 8 } \int \frac { d ( 1 + x ^ { 8 } ) } { 1+ x ^ { 8 } } $ $\displaystyle \ = \frac { 1 } { 8 } \ln ( 1 + x ^ { 8 } ) + C $

$ (3) \quad $When $k=1$ , $\displaystyle \quad I _ { 1 } = \int \frac { x } { 1 + x ^ { 8 } } d x = \frac { 1 } { 2 } \int \frac { d ( x ^ { 2 } ) } { 1 + ( x ^ { 2 } ) ^ { 4 } }= \frac { 1 } { 2 } \int \frac { d y } { 1 + y ^ { 4 } } , \text { where } y = x ^ { 2}.$

$$\int \frac{x}{1+x^{8}} d x =\frac{1}{4 \sqrt{2}} \tan ^{-1}\left(\frac{x^{4}-1}{\sqrt{2} x^{2}}\right)+\frac{1}{8 \sqrt{2}} \ln \left| \frac{x^{4}+\sqrt{2} x^{2}+1}{x^{4}-\sqrt{2} x^{2}+1}\right|+C$$

For detailed solution, please refer to my post.

Now we are going to tackle $I_2, I_4, I_5 $ and $ I_6 $ using a wonderful identity $(*)$.

$$\displaystyle \because 1 + x ^ { 8 } \displaystyle = ( x ^ { 4 } + 1 ) ^ { 2 } - 2 x ^ { 4 }\displaystyle = \left( x ^ { 4 } + 1 ) ^ { 2 } - ( \sqrt { 2 } x ^ { 2 } \right) ^ { 2 }\displaystyle = \left( x ^ { 4 } + \sqrt { 2 } x ^ { 2 } + 1 ) ( x ^ { 4 } - \sqrt { 2 } x ^ { 2 } + 1 \right)$$ $$\displaystyle \therefore \frac { x ^ { 2 } } { 1 + x ^ { 8 } } = \frac { 1 } { 2 \sqrt { 2 } }\left( \frac { 1 } { x ^ { 4 } - \sqrt { 2 } x ^ { 2 } + 1 } \cdot \frac { 1 } { x ^ { 4 } + \sqrt { 2 } x ^ { 2 } + 1 } \right) $$

$\displaystyle I _ { 2 } = \int \frac { x ^ { 2 } } { 1 + x ^ { 8 } } d x \displaystyle \ = \frac { 1 } { 2 \sqrt { 2 } } \int \left( \frac { 1 } { x ^ { 4 } - \sqrt { 2 } x ^ { 2 } + 1 } - \frac { 1 } { x ^ { 4 } + \sqrt { 2 } x ^ { 2 } + 1 } \right) d x $

$\displaystyle \quad = \frac { 1 } { 4 \sqrt {4-2 \sqrt{2}}} \tan ^ { - 1 } \left( \frac { x ^ { 2 } - 1 } { \sqrt { 2 - \sqrt { 2 } } x } \right)$ $ \displaystyle +\frac { 1 } { 8 \sqrt {4+2\sqrt{2}}} \ln \left| \frac { x ^ { 2 } + \sqrt { 2 + \sqrt { 2 } } x + 1 } { x ^ { 2 } - \sqrt { 2 + \sqrt { 2 } } x + 1 } \right|$ $ \qquad \displaystyle - \frac { 1 } { 4 \sqrt { 4+ 2 \sqrt{2} } } \tan ^ { - 1 } \left( \frac { x ^ { 2 } - 1 } { \sqrt { 2 + \sqrt { 2 } } x } \right)$ $\displaystyle - \frac { 1 } { 8 \sqrt { 4-2 \sqrt { 2 }} } \ln \left| \frac { x ^ { 2 } + \sqrt { 2 - \sqrt { 2 } } x + 1 } { x ^ { 2 } - \sqrt { 2 - \sqrt { 2 } } x + 1 } \right| + C$

For detailed solution, please refer to my post

$\displaystyle I _ { 4 } = \int \frac { x ^ { 4 } } { 1 + x ^ { 8 } } d x $ $\\ \displaystyle = \int x ^ { 2 } \cdot \frac { x ^ { 2 } } { 1 + x ^ { 8 } } d x $ $\displaystyle = \int\left( \frac { x ^ { 2 } } { x ^ { 4 } - \sqrt { 2 } x ^ { 2 } + 1 } - \frac { x ^ { 2 } } { x ^ { 4 } + \sqrt { 2 } x ^ { 2 } + 1 } \right) d x $

$ \displaystyle =\frac{1}{4 \sqrt{2}}\left[\frac{1}{2 \sqrt{2+\sqrt{2}}} \ln \left|\frac{x^{2}-\sqrt{2+\sqrt{2}} x+1}{x^{2}+\sqrt{2+\sqrt{2}} x+1}\right|+\frac{1}{\sqrt{2-\sqrt{2}}} \tan^{-1}\left(\frac{x^{2}-1}{\sqrt{2-\sqrt{2}} x}\right)\\ \quad - \frac{1}{2 \sqrt{2-\sqrt{2}}} \ln \left|\frac{x^{2}-\sqrt{2-\sqrt{2}} x+1}{x^{2}+\sqrt{2-\sqrt{2} x}+1}\right|-\frac{1}{\sqrt{2+\sqrt{2}}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{2+\sqrt{2}} x}\right)\right]+C$

For detailed solution, please refer to my post

$$\displaystyle I _ { 5 } = \int x ^ { 3 } \cdot \frac { x ^ { 2 } } { 1 + x ^ { 8 } } d x \displaystyle = \int \left( \frac { x ^ { 3 } } { x ^ { 4 } - \sqrt { 2 } x ^ { 2 } + 1 } - \frac { x ^ { 3 } } { x ^ { 4 } + \sqrt { 2 } x ^ { 2 } + 1 }\right) dx$$

$$=\frac{1}{8\sqrt{2}} \ln \left|\frac{x^{4}-\sqrt{2} x^{2}+1}{x^{4}+\sqrt{2} x^{2}+ 1}\right|+\frac{1}{4\sqrt{2}} \tan ^{-1}\left(\frac{\sqrt{2} x^{2}}{1-x^4}\right)+C$$

For detailed solution, please refer to my post

$$\left. \begin{array}{l}{ \displaystyle I _ { 6 } = \int x ^ { 4 } \cdot \frac { x ^ { 2 } } { 1 + x ^ { 8 } } d x }\\{ \displaystyle \quad = \int \left( \frac { x ^ { 4 } } { x ^ { 4 } - \sqrt { 2 } x ^ { 2 } + 1 } - \frac { x ^ { 4 } } { x ^ { 4 } + \sqrt { 2 } x ^ { 2 } + 1 } \right) d x }\\{ \displaystyle = \int\left [\left( 1 + \frac { \sqrt { 2 } x - 1 } { x ^ { 4 } - \sqrt { 2 } x ^ { 2 } + 1 } \right) - \left( 1 - \frac { \sqrt { 2 } x + 1 } { x ^ { 4 } + \sqrt { 2 } x ^ { 2 } + 1 } \right)\right] d x }\\{ \displaystyle = \int \left( \frac { \sqrt { 2 } x - 1 } { x ^ { 4 } - \sqrt { 2 } x ^ { 2 } + 1 } - \frac { \sqrt { 2 } x + 1 } { x ^ { 4 } + \sqrt { 2 } x ^ { 2 } + 1 } \right) d x }\end{array} \right.$$ $\displaystyle =\frac{\sqrt{2}+1}{8 \sqrt{4+2\sqrt{2}}}\ln \left|\frac{x^{2}-\sqrt{2+\sqrt{2}} x+1}{x^{2}+\sqrt{2+\sqrt{2} x+1}}\right| \\$ $\displaystyle +\frac{\sqrt{2}-1}{4 \sqrt{4-\sqrt{2}} } \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{2-\sqrt{2}}x}\right)$ $ \displaystyle +\frac{\sqrt{2}-1}{8 \sqrt{4-\sqrt{2}} }\ln \left| \frac{x^{2}-\sqrt{2-\sqrt{2}} x+1}{x^{2}+\sqrt{2-\sqrt{2}} x+1} \right|\\$ $\displaystyle \quad +\frac{\sqrt{2}+1}{4 \sqrt{4+2\sqrt{2}}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{2+\sqrt{2}}x}\right)+C$

For detailed solution, please refer to my post

I have just finished the integrals $I_1-I_7 $ but not $I_0. $

After thinking for hours, I finally found a way to tackle $I_0$ by dividing (*) by $x^2$,

$$\displaystyle \int \frac { 1 } { 1 + x ^ { 8 } } d x = \frac { 1 } { 2 \sqrt { 2 } } \left[ \int \frac { 1 } { x ^ { 2 } ( x ^ { 4 } - \sqrt { 2 } x ^ { 2 } + 1 ) } - \frac { 1 } { x ^ { 2 } ( x ^ { 4 } + \sqrt { 2 } x ^ { 2 } + 1 ) }\right ]$$

$$\displaystyle \because \frac { 1 } { x ^ { 2 } ( x ^ { 4 } - \sqrt { 2 } x ^ { 2 } + 1 ) } = \frac { 1 } { x ^ { 2 } } - \frac { x ^ { 2 } - \sqrt { 2 } } { x ^ { 4 } - \sqrt { 2 } x ^ { 2 } + 1 } \displaystyle \textrm{and } \frac { 1 } { x ^ { 2 } ( x ^ { 4 } + \sqrt { 2 } x ^ { 2 } + 1 ) } = \frac { 1 } { x ^ { 2 } } - \frac { x ^ { 2 } + \sqrt { 2 } } { x ^ { 4 } + \sqrt { 2 } x ^ { 2 } + 1 }$$

$$\displaystyle \therefore \int \frac { 1 } { 1 + x ^ { 8 } } d x \displaystyle = \frac { 1 } { 2 \sqrt { 2 } } \int \left( \frac { x ^ { 2 } + \sqrt { 2 } } { x ^ { 4 } + \sqrt { 2 } x ^ { 2 } + 1 } - \frac { x ^ { 2 } - \sqrt { 2 } } { x ^ { 4 } -\sqrt { 2 } x ^ { 2 } + 1 } \right) d x $$ $\displaystyle \left. \begin{array} { l } {\displaystyle = \frac { \sqrt { 2 } + 1 } { 4 \sqrt { 4 + 2 \sqrt { 2 } } } \tan ^ { - 1 }\left( \frac { x ^ { 2 } - 1 } { \sqrt { 2 + \sqrt { 2 } } x } \right) +\frac { \sqrt { 2 } - 1 } { 8 \sqrt { 4 - 2 \sqrt { 2 } } } \ln \left| \frac { x ^ { 2 } + \sqrt { 2 - \sqrt { 2 } } x + 1 } { x ^ { 2 } - \sqrt { 2 - \sqrt { 2 } x + 1 } } \right|} \\ { \displaystyle + \frac { \sqrt { 2 } - 1 } { 4 \sqrt { 4 - 2 \sqrt { 2 } } } \tan ^ { - 1 } \left( \frac { x ^ { 2 } - 1 } { \sqrt { 2 - \sqrt { 2 } } x } \right) + \frac { \sqrt { 2 } + 1 } { 8 \sqrt { 4 + 2 \sqrt { 2 } } } \ln \left| \frac { x ^ { 2 }+\sqrt { 2 + \sqrt { 2 }} x + 1 } { x ^ { 2 } - \sqrt { 2 + \sqrt { 2 } } x + 1 } \right| + C }\end{array} \right.$

For detailed solution, please refer to my post

By long division, we can evaluate all the integrals in the form $\displaystyle \int \frac{P(x)}{1+x^8} dx,\tag*{} $ $ \textrm{ where }P(x)\textrm{ is a polynomial in }x.$

Wish you enjoy the solutions!

My Question Is there a faster method to find the integrals?

Lai
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    With the $k = 1$ case in hand, you can resolve $k = 5$ using the substitution $$\int \frac{x^5 ,dx}{1 + x^8} \stackrel{u = \frac{1}{x}}{=} -\int \frac{u ,du}{1 + u^8} .$$ – Travis Willse Mar 16 '22 at 05:54
  • @TravisWillse You know, that makes perfect sense to me, but if we plug $k=3$ in doesn't the same math say that $\int\dfrac{x^3}{x^8+1}dx = -\int\dfrac{u^3}{u^8+1}du$? – Steven Stadnicki Mar 16 '22 at 06:02
  • @StevenStadnicki I think we're okay: The two sides are equal, up to addition of a constant: Back-substituting gives $-\frac{1}{4} \arctan(u^4) + C = -\frac{1}{4} \arctan \left(\frac{1}{x^4}\right) + C$. But $\arctan y = \frac{\pi}{2} - \arctan \frac{1}{y}$ (on $(0,\infty)$), so the two expressions agree. – Travis Willse Mar 16 '22 at 06:30
  • @Lai The explicit antiderivatives for even $k$ are pretty messy, so there's probably no getting around some tedious manipulation entirely. But the same substitution essentially reduces the $k = 4$ and $k = 6$ cases to the $k = 2$ and $k = 0$ cases, respectively. Incidentally, the definite integrals of these functions over $\Bbb R$ or $[0, \infty)$ are a nice application of contour integrals, and you can handle them all simultaneously by leaving $k$ unspecified and picking an appropriate contour for each value of $k$. – Travis Willse Mar 16 '22 at 06:37
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    Are you familiar with complex elementary functions (logarithms, trig functions and such)? You can do partial fractions expansions to any of these like $$\frac{x^k}{1+x^8}=\sum_{j=0}^7\frac{C_{k,j}}{1+\zeta^{2j+}x},$$ where $\zeta=e^{\pi i/8}$ is a primitive sixteenth root of unity, and $C_{k,j}$ are complex constants. Then integrate the terms using complex logarithms. It won't be pretty, but not too bad :-) – Jyrki Lahtonen Mar 16 '22 at 07:29
  • https://people.math.umass.edu/~cattani/hypergeom_lectures.pdf https://webspace.science.uu.nl/~beuke106/AHGcourse.pdf – Claude Leibovici Mar 16 '22 at 09:36
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    @TravisWillse Ahh, of course — I forgot that after the substitution $u$ isn't exactly a 'formal variable', so we can't use this to say that an antiderivative $F$ of $\dfrac{x^3}{x^8+1}$ satisfies $F(t)=C-F(t)$. (Or equivalently, the variable limits for the indefinite integration have to be changed along with the substitution) – Steven Stadnicki Mar 16 '22 at 18:40
  • @Lai Cf. https://math.stackexchange.com/questions/2409312/evaluating-int-fracxb1xadx-for-a-b-in-bbb-n – Travis Willse Mar 16 '22 at 23:46

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Once more, the gaussian hypergeometric function $$\int \frac{x^k}{1+x^p}\,dx=\frac{x^{k+1}}{k+1}\,\,\, _2F_1\left(1,\frac{k+1}{p};\frac{k+p+1}{p};-x^p\right)$$ $$\int_0^\infty \frac{x^k}{1+x^p}\,dx=\frac \pi p \,\csc \left(\frac{k+1}{p}\pi\right)$$ $$\int_0^1 \frac{x^k}{1+x^p}\,dx=\frac 1{2p} \left(\psi \left(\frac{k+p+1}{2 p}\right)-\psi \left(\frac{k+1}{2 p}\right)\right)$$

Claude Leibovici
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  • Thank you for your fast solution! How can I start with hypergeometric function? Is there any good book or reference to start with? – Lai Mar 16 '22 at 07:28
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    @Lai. Google for "books about hypergeometric functions for beginners". You will find many. This one is free http://www.fuchs-braun.com/media/e4ef05831b027c9cffff8033ffffffef.pdf – Claude Leibovici Mar 16 '22 at 07:37
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I'll leave you with this formula that I came up with a few years ago:

$$\frac{x^m}{x^n+a^n}=a^{m-n}\left[\frac{a}{n}\sum_{k=1}^{n}\frac{\cos\left(\frac{2k+1}{n}m\pi\right)a-\cos\left(\frac{2k+1}{n}(m+1)\pi\right)x}{x^2-2\cos\left(\frac{2k+1}{n}\pi\right)ax+a^2}-\sum_{k=1}^{\left\lfloor\frac{m}{n}\right\rfloor}(-1)^k\left(\frac{x}{a}\right)^{m-nk}\right]$$

The integration is very easy but takes a long time to write (essentially you get logarithms, arctangents and polynomials).

However, in your case with $a=1, n=8$ and $m<n$ it is easier:

$$\frac{x^m}{x^8+1}=\frac{1}{8}\sum_{k=1}^{8}\frac{\cos\left(\frac{2k+1}{8}m\pi\right)-\cos\left(\frac{2k+1}{8}(m+1)\pi\right)x}{x^2-2\cos\left(\frac{2k+1}{8}\pi\right)x+1}\qquad \text{for }m=1,...,7$$

Math Attack
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