Question

Найти неопределенный интеграл
$\frac{dx}{x^4\sqrt{x^2-1} }$

Answer 1

Ответ:

$\int \dfrac{dx}{x^4\sqrt{x^2-1}}=\Big[\ x=\dfrac{1}{cost}\ ,\ dx=\dfrac{sint\, dt}{cos^2t}\ ,\ x^2-1=\dfrac{1}{cos^2t}-1=tg^2t\ \Big]=\\\\\\=\int \dfrac{sint\, dt}{cos^2t\cdot tgt}=\int \dfrac{sint\cdot cost\, dt}{cos^2t\cdot \dfrac{1}{cos^4t}\cdot sint}=\int cos^3t\, dt=\int cos^2t\cdot cost\, dt=\\\\\\=\int (1-sin^2t)\cdot cost\, dt=\Big[\ u=sint\ ,\ du=cost\, dt\ \Big]=\int (1-u^2)\, du=\\\\\\=\int du-\int u^2\, du=u-\dfrac{u^3}{3}+C=sint-\dfrac{sin^3t}{3}+C=$

$=sin\Big(arccos\dfrac{1}{x}\Big)-\dfrac{sin^3(arccos\frac{1}{x})}{3}+C=\dfrac{\sqrt{x^2-1}}{x}-\dfrac{\sqrt{(x^2-1)^3}}{3\, x^3}+C$