Question

Неопределенный интеграл
1/(x+sqrt(x^2-1))^2

Answer 1

Ответ:

$\displaystyle\frac{1-3(x+\sqrt{x^2-1})^2}{6(x+\sqrt{x^2-1})}+const$

Пошаговое объяснение:

$\displaystyle\int {\frac1{(x+\sqrt{x^2-1})^2}} \, dx =\\\left\{{x=\frac{u^2+1}{2u}\atop{dx=(1-\frac{u^2+1}{2u^2})}} \right \} =\\\int{\frac{u^2-1}{2u^4}}\,du=\\н\int\frac1{u^2}-\frac1{u^4}\,du=\\н\int\frac1{u^2}\,du-н\int\frac1{u^4}\,du=\\-\frac1{2u}+\frac1{6u^3}+const=\\\left\{u=x+\sqrt{x^2-1}\right\}=\\\frac{\frac1{(x+\sqrt{x^2-1})^2}-3}{6(x+\sqrt{x^2-1})}+const=\\\frac{1-3(x+\sqrt{x^2-1})^2}{6(x+\sqrt{x^2-1})}+const$