$$\int \dfrac{x^3\,dx}{x^8-2}$$

Так как $d(x^4)=4x^3\,dx$, то $x^3\,dx=\dfrac14\,d(x^4)$, а $x^8=(x^4)^2$: $$\int\frac{x^3\,dx}{x^8-2}=\frac14\int\frac{d(x^4)}{(x^4)^2-2}.$$ По табличной формуле $\displaystyle\int\frac{du}{u^2-a^2}=\frac1{2a}\ln\left|\frac{u-a}{u+a}\right|+C$ при $u=x^4$, $a=\sqrt2$: $$\frac14\cdot\frac1{2\sqrt2}\ln\left|\frac{x^4-\sqrt2}{x^4+\sqrt2}\right|+C=\frac1{8\sqrt2}\ln\left|\frac{x^4-\sqrt2}{x^4+\sqrt2}\right|+C.$$

$$\boxed{\dfrac1{8\sqrt2}\ln\left|\dfrac{x^4-\sqrt2}{x^4+\sqrt2}\right|+C}$$