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