What is the solution for f^'x=(2x)(x^4-8x)+(x^2)(4x^3-8) ?

The solution for f^'x=(2x)(x^4-8x)+(x^2)(4x^3-8) is f(x)=6((x^5)/5-2x^2)+c_{1}

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f^'x=(2x)(x^4-8x)+(x^2)(4x^3-8)

f^{^{'}} x = (2 x) (x^{4} - 8 x) + (x^{2}) (4 x^{3} - 8)

f (x) = 6 (\frac{x ^{5}}{5} - 2 x^{2}) + c_{1}

Solution steps

f^{'} (x) x = (2 x) (x^{4} - 8 x) + (x^{2}) (4 x^{3} - 8)

Isolate

f^{'} (x) : f^{'} (x) = 6 x (x^{3} - 4)

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

then

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

f (x) = \int 6 x (x^{3} - 4) d x

\int 6 x (x^{3} - 4) d x = 6 (\frac{x ^{5}}{5} - 2 x^{2}) + c_{1}

f (x) = 6 (\frac{x ^{5}}{5} - 2 x^{2}) + c_{1}

y = \frac{6 t - 3}{5 t + 8}

x \to \frac{π}{2} lim (sin (\frac{x}{2}))

n = 3 \sum \infty (\frac{e}{π})^{n}

\frac{1}{x ^{2} + 1}

\frac{1}{( s ^{2} + 4 ) ( s ^{2} + 1 )}

What is the solution for f^'x=(2x)(x^4-8x)+(x^2)(4x^3-8) ?
The solution for f^'x=(2x)(x^4-8x)+(x^2)(4x^3-8) is f(x)=6((x^5)/5-2x^2)+c_{1}