Задача №1554

\[ \int\frac{\sqrt{1+x^8}}{x^{13}}dx =\int\frac{x^3\sqrt{1+x^8}}{x^{16}}dx =\left[u=x^4\right] =\frac{1}{4}\cdot\int\frac{\sqrt{1+u^2}}{u^4}du=\\ =-\frac{1}{4}\cdot\int\frac{\sqrt{1+u^2}}{u^2}d\left(\frac{1}{u}\right) =\left[t=\frac{1}{u}\right] =-\frac{1}{4}\cdot\int{t^2}\sqrt{1+\frac{1}{t^2}}dt =-\frac{1}{8}\cdot\int\left(t^2+1\right)^{\frac{3}{2}}d\left(t^2+1\right)=\\ =-\frac{1}{12}\cdot\left(t^2+1\right)^{\frac{3}{2}}+C =-\frac{1}{12}\cdot\left(\frac{1}{u^2}+1\right)^{\frac{3}{2}}+C =-\frac{1}{12}\cdot\left(\frac{1}{x^8}+1\right)^{\frac{3}{2}}+C =-\frac{\sqrt{\left(x^8+1\right)^3}}{12x^{12}}+C. \]