What is the solution for (dy)/(dx)= 1/(x^4+4x^2) ?

The solution for (dy)/(dx)= 1/(x^4+4x^2) is y=-1/(4x)-1/8 arctan(x/2)+c_{1}

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(dy)/(dx)= 1/(x^4+4x^2)

\frac{d y}{d x} = \frac{1}{x ^{4} + 4 x ^{2}}

y = - \frac{1}{4 x} - \frac{1}{8} arctan (\frac{x}{2}) + c_{1}

Solution steps

\frac{d y}{d x} = \frac{1}{x ^{4} + 4 x ^{2}}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = \frac{1}{x ^{4} + 4 x ^{2}}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{1}{x ^{4} + 4 x ^{2}} d x

\int \frac{1}{x ^{4} + 4 x ^{2}} d x = - \frac{1}{4 x} - \frac{1}{8} arctan (\frac{x}{2}) + c_{1}

y = - \frac{1}{4 x} - \frac{1}{8} arctan (\frac{x}{2}) + c_{1}

x \to 2 lim (sin (\frac{1}{x - 2}))

x \to \infty lim (\frac{4 x ^{2} + 5 x}{10 + x ^{4}})

3 y^{^{''}} - 12 y = 0

x \to 0 + lim (\frac{1}{x ^{3} - 2 x ^{2}})

y = x^{4} + 5 x^{2} - x

What is the solution for (dy)/(dx)= 1/(x^4+4x^2) ?
The solution for (dy)/(dx)= 1/(x^4+4x^2) is y=-1/(4x)-1/8 arctan(x/2)+c_{1}