What is the solution for f^'(x)= 3/4*(1-x^2) ?

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

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f^'(x)= 3/4*(1-x^2)

f^{^{'}} (x) = \frac{3}{4} \cdot (1 - x^{2})

f (x) = \frac{3}{4} x - \frac{x ^{3}}{4} + c_{1}

Solution steps

f^{'} (x) = \frac{3}{4} \cdot (1 - x^{2})

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

then

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

f (x) = \int \frac{3}{4} (1 - x^{2}) d x

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

f (x) = \frac{3}{4} x - \frac{x ^{3}}{4} + c_{1}

\frac{d}{d x} (x^{2} + 6 x + 3)

\frac{d}{d x} (3 x^{2} - 12 x + 9)

11 t^{2} y^{^{''}} - 22 y = 9 t^{2} - 9

x \to 0 lim (cos (2 - 2 e^{x}))

x \to 1 lim (2)

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