Задача №1554
Условие
Найти интеграл \(\int\frac{\sqrt{1+x^8}}{x^{13}}dx\).
Решение
\[
\int\frac{\sqrt{1+x^8}}{x^{13}}dx
=\int\frac{x^3\sqrt{1+x^8}}{x^{16}}dx
=\left[u=x^4\right]
=\frac{1}{4}\cdot\int\frac{\sqrt{1+u^2}}{u^4}du=\\
=-\frac{1}{4}\cdot\int\frac{\sqrt{1+u^2}}{u^2}d\left(\frac{1}{u}\right)
=\left[t=\frac{1}{u}\right]
=-\frac{1}{4}\cdot\int{t^2}\sqrt{1+\frac{1}{t^2}}dt
=-\frac{1}{8}\cdot\int\left(t^2+1\right)^{\frac{3}{2}}d\left(t^2+1\right)=\\
=-\frac{1}{12}\cdot\left(t^2+1\right)^{\frac{3}{2}}+C
=-\frac{1}{12}\cdot\left(\frac{1}{u^2}+1\right)^{\frac{3}{2}}+C
=-\frac{1}{12}\cdot\left(\frac{1}{x^8}+1\right)^{\frac{3}{2}}+C
=-\frac{\sqrt{\left(x^8+1\right)^3}}{12x^{12}}+C.
\]
Ответ:
\(-\frac{\sqrt{\left(x^8+1\right)^3}}{12x^{12}}+C\)