Интеграл x^3/(x^2-1)dx срочно

Решите задачу:

$\int{ \frac{ x^3 }{ x^2 - 1 } } \, dx = \frac{1}{2} \int{ \frac{ x^2 }{ x^2 - 1 } } \, 2xdx \ = \frac{1}{2} \int{ \frac{ x^2 - 1 + 1 }{ x^2 - 1 } } \, dx^2 = \\\\ = \frac{1}{2} ( \int{ \frac{ x^2 - 1 }{ x^2 - 1 } } \, dx^2 + \int{ \frac{1}{ x^2 - 1 } } \, dx^2 ) = \frac{1}{2} ( \int{ dx^2 } + \int{ \frac{ d( x^2 - 1 ) }{ x^2 - 1 } } ) = \\\\ = \frac{1}{2} ( x^2 + \ln{ | x^2 - 1 | } ) + C = \frac{ x^2 + \ln{ | x^2 - 1 | } }{2} + C \ ;$