What is the solution for f^'(x)=7x^4-9x^3-4 ?

The solution for f^'(x)=7x^4-9x^3-4 is f(x)=(7x^5)/5-(9x^4)/4-4x+c_{1}

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f^'(x)=7x^4-9x^3-4

f^{^{'}} (x) = 7 x^{4} - 9 x^{3} - 4

f (x) = \frac{7 x ^{5}}{5} - \frac{9 x ^{4}}{4} - 4 x + c_{1}

Solution steps

f^{'} (x) = 7 x^{4} - 9 x^{3} - 4

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

then

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

f (x) = \int 7 x^{4} - 9 x^{3} - 4 d x

\int 7 x^{4} - 9 x^{3} - 4 d x = \frac{7 x ^{5}}{5} - \frac{9 x ^{4}}{4} - 4 x + c_{1}

f (x) = \frac{7 x ^{5}}{5} - \frac{9 x ^{4}}{4} - 4 x + c_{1}

\frac{\partial}{\partial x} (\frac{x}{x - b})

\int x (sec (x) tan (x)) d x

4 x^{3} - 6 x^{2}

\int_{5}^{\infty} \frac{1}{x ^{2}} d x

\frac{d}{d x} (tan^{2} (x - 3))

What is the solution for f^'(x)=7x^4-9x^3-4 ?
The solution for f^'(x)=7x^4-9x^3-4 is f(x)=(7x^5)/5-(9x^4)/4-4x+c_{1}