Интеграл (x^4-2x^2-1)/(2x^2+1) dx срочно пожалуйста

Решите задачу:

$\int{ \frac{ x^4 - 2x^2 - 1 }{ 2x^2 + 1 } } \, dx = \frac{1}{4} \int{ \frac{ 4x^4 - 8x^2 - 4 }{ 2x^2 + 1 } } \, dx = \frac{1}{4} \int{ \frac{ 4x^4 + 4x^2 + 1 - 12x^2 - 5 }{ 2x^2 + 1 } } \, dx = \\\\ = \frac{1}{4} \int{ \frac{ ( 2x^2 + 1 )^2 - 12x^2 - 5 }{ 2x^2 + 1 } } \, dx = \frac{1}{4} ( \int{ \frac{ ( 2x^2 + 1 )^2 }{ 2x^2 + 1 } } \, dx - \int{ \frac{ 12x^2 + 5 }{ 2x^2 + 1 } } \, dx ) = \\\\ = \frac{1}{4} ( \int{ \frac{ ( 2x^2 + 1 )^2 }{ 2x^2 + 1 } } \, dx - \int{ \frac{ 6 ( 2x^2 + 1 ) - 1 }{ 2x^2 + 1 } } \, dx ) = \\\\ = \frac{1}{4} ( \int{ ( 2x^2 + 1 ) } \, dx - \int{ \frac{ 6 ( 2x^2 + 1 ) }{ 2x^2 + 1 } } \, dx + \int{ \frac{1}{ 2x^2 + 1 } } \, dx ) = \\\\ = \frac{1}{2} \int{x^2} \, dx + \frac{1}{4} \int{dx} - \frac{3}{2} \int{dx} + \frac{1}{ 4 \sqrt{2} } \int{ \frac{ d( \sqrt{2} x ) }{ ( \sqrt{2} x )^2 + 1 } } = \\\\ = \frac{1}{6} x^3 - \frac{5}{4} \int{dx} + \frac{1}{ 4 \sqrt{2} } arctg{ \sqrt{2} x } = \frac{ x^3 }{6} - \frac{5}{4} x + \frac{ arctg{ \sqrt{2} x } }{ 4 \sqrt{2} } + C \ ;$

$\int{ \frac{ x^4 - 2x^2 - 1 }{ 2x^2 + 1 } } \, dx = \frac{ x^3 }{6} - \frac{5}{4} x + \frac{ arctg{ \sqrt{2} x } }{ 4 \sqrt{2} } + C \ ;$