Đáp án:
$\begin{array}{l}
y = \frac{x}{{\sqrt[8]{{{x^8} + 1}}}} = x.{\left( {{x^8} + 1} \right)^{\frac{{ - 1}}{8}}}\\
\Rightarrow y' = {\left( {{x^8} + 1} \right)^{ - \frac{1}{8}}} + x.\left( { - \frac{1}{8}} \right).\left( {{x^8} + 1} \right)'.{\left( {{x^8} + 1} \right)^{\frac{{ - 1}}{8} - 1}}\\
= \frac{1}{{\sqrt[8]{{{x^8} + 1}}}} - x.\frac{1}{8}.8{x^7}.\frac{1}{{\left( {{x^8} + 1} \right).\sqrt[8]{{{x^8} + 1}}}}\\
= \frac{1}{{\sqrt[8]{{{x^8} + 1}}}} - \frac{{{x^8}}}{{\left( {{x^8} + 1} \right).\sqrt[8]{{{x^8} + 1}}}}\\
= \frac{{{x^8} + 1 - {x^8}}}{{\left( {{x^8} + 1} \right).\sqrt[8]{{{x^8} + 1}}}}\\
= \frac{1}{{\left( {{x^8} + 1} \right).\sqrt[8]{{{x^8} + 1}}}}\\
\Rightarrow \int {\frac{1}{{\left( {{x^8} + 1} \right).\sqrt[8]{{{x^8} + 1}}}}dx} = \frac{x}{{\sqrt[8]{{{x^8} + 1}}}}
\end{array}$