Break up the right way

$Ques.\quad \int { \frac { dx }{ { x }^{ 6 }+{ x }^{ 4 } } } \\ =\int { \frac { dx }{ { x }^{ 4 }({ x }^{ 2 }+1) } } \\ =\int { \frac { dx({ x }^{ 2 }+1-{ x }^{ 2 }) }{ { x }^{ 4 }({ x }^{ 2 }+1) } } \\ =\int { \left[ \frac { 1 }{ { x }^{ 4 } } -\frac { 1 }{ { x }^{ 2 }({ x }^{ 2 }+1) } \right] } \\ =\int { \left[ \frac { 1 }{ { x }^{ 4 } } -\frac { { x }^{ 2 }+1-{ x }^{ 2 } }{ { x }^{ 2 }({ x }^{ 2 }+1) } \right] } \\ =\int { \left[ \frac { 1 }{ { x }^{ 4 } } -\frac { 1 }{ { x }^{ 2 }) } +\frac { 1 }{ 1+{ x }^{ 2 } } \right] } \\ \Rightarrow \frac { -1 }{ 3{ x }^{ 3 } } +\frac { 1 }{ x } +\arctan { x }$

When you apply partial fractions , you get , $\int \frac{dx}{x^6+x^4} = \int [-\frac{1}{x^2} + \frac{1}{x^4} + \frac{1}{x^2+1}]dx$ The integration of which is, $\int \frac{-1}{x^2} = \frac{1}{x} , \int \frac{1}{x^4} = \frac{-1}{3x^3},\int \frac{1}{x^2+1} = tan^{-1}x / arctan x$ So we get, $\int \frac{dx}{x^6+x^4} =\displaystyle \boxed{ \dfrac {-1}{3x^3} + \dfrac{1}{x} +arctan x + c}$

Neste problema devemos aplicar a técnica de integração por partes. Inicialmente, devemos trabalhar com o integrando. Segue,

$\\ \frac{1}{x^6 + x^4} = \frac{1}{x^4(x^2 + 1)} \\\\\\ \frac{1}{x^6 + x^4} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x^4} + \frac{Ex + F}{x^2 + 1} \\\\\\ \frac{1}{x^6 + x^4} = \frac{A(x^2 + 1)x^3 + B(x^2 + 1)x^2 + C(x^2 + 1)x + D(x^2 + 1) + (Ex + F)x^4}{x^4(x^2 + 1)}$

Resolvendo o sistema formado obtemos: $\boxed{A = 0}, \; \boxed{B = - 1}, \; \boxed{C = 0}, \; \boxed{D = 1}, \; \boxed{E = 0}, \; \boxed{F = 1}$ .

Portanto,

$\int \frac{dx}{x^6 + x^4} = \int \frac{- 1}{x^2} \, dx + \int \frac{1}{x^4} \, dx + \int \frac{1}{x^2 + 1} \, dx \\\\\\ \int \frac{dx}{x^6 + x^4} = (x^{- 1} + c_1) + (- \frac{x^{- 3}}{3} + c_2) + (\tan^{- 1} x + c_3) \\\\\\ \boxed{\boxed{\int \frac{dx}{x^6 + x^4} = \frac{1}{x} - \frac{1}{3x^3} + \tan^{- 1} x + C}}$