Задача №1555
Условие
Найти интеграл \(\int\frac{xdx}{\left(1-x^4\right)^{\frac{3}{2}}}\).
Решение
Первый способ
\[
\int\frac{xdx}{\left(1-x^4\right)^{\frac{3}{2}}}
=\left[t=x^2\right]
=\frac{1}{2}\int\frac{dt}{\left(1-t^2\right)^{\frac{3}{2}}}
=\left[\begin{aligned}
& t=\sin{z};\;t\in\left[0;\frac{\pi}{2}\right);\\
& dt=\cos{z}dz.
\end{aligned}\right]=\\
=\frac{1}{2}\int\frac{dz}{\cos^2{z}}
=\frac{\tg{z}}{2}+C
=\frac{\sin{z}}{2\sqrt{1-\sin^2{z}}}+C
=\frac{t}{2\sqrt{1-t^2}}+C
=\frac{x^2}{2\sqrt{1-x^4}}+C.
\]
Второй способ
\[
\int\frac{xdx}{\left(1-x^4\right)^{\frac{3}{2}}}
=\int\frac{x^3dx}{x^2\left(1-x^4\right)^{\frac{3}{2}}}
=-\frac{1}{4}\int\frac{d\left(1-x^4\right)}{x^2\cdot\left(1-x^4\right)^{\frac{3}{2}}}
=\left[t=1-x^4\right]=\\
=-\frac{1}{4}\int\frac{dt}{\sqrt{1-t}\cdot{t^{\frac{3}{2}}}}
=-\frac{1}{4}\int\frac{dt}{t^2\sqrt{\frac{1}{t}-1}}
=\frac{1}{4}\int\frac{d\left(\frac{1}{t}-1\right)}{\sqrt{\frac{1}{t}-1}}
=\frac{1}{2}\cdot\sqrt{\frac{1}{t}-1}
=\frac{x^2}{2\sqrt{1-x^4}}+C.
\]
Ответ:
\(\frac{x^2}{2\sqrt{1-x^4}}+C.\)