Giả sử $u=\sqrt{x}$. Thế thì $x=u^2$, và do đó $dx=2udu$.Từ đó $I=\int\limits_{\frac{1}{\sqrt{2}}}^{1}\frac{u}{\sqrt{u^6+1}}2udu$
$=2\int\limits_{\frac{1}{\sqrt{2}}}^{1}\frac{u^2}{\sqrt{u^6+1}}du$
$=\frac{2}{3}\int\limits_{\frac{1}{\sqrt{2}}}^{1}\frac{1}{\sqrt{(u^3)^2+1}}d(u^3)$
$=\frac{2}{3}(ln|u^3+\sqrt{u^6+1}|)|^{1}_{\frac{1}{\sqrt{2}}}$
$=\frac{2}{3}ln(\frac{2+\sqrt{2}}{2})$.