What is the solution for f^'(x)=(x^2)/((9+4x)) ?

The solution for f^'(x)=(x^2)/((9+4x)) is f(x)=(x^2)/8-9/16 x+81/64 ln(4x+9)+c_{1}

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f^'(x)=(x^2)/((9+4x))

f^{^{'}} (x) = \frac{x ^{2}}{( 9 + 4 x )}

f (x) = \frac{x ^{2}}{8} - \frac{9}{16} x + \frac{81}{64} ln (4 x + 9) + c_{1}

Solution steps

f^{'} (x) = \frac{x ^{2}}{( 9 + 4 x )}

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

then

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

f (x) = \int \frac{x ^{2}}{9 + 4 x} d x

\int \frac{x ^{2}}{9 + 4 x} d x = \frac{x ^{2}}{8} - \frac{9}{16} x + \frac{81}{64} ln (4 x + 9) + c_{1}

f (x) = \frac{x ^{2}}{8} - \frac{9}{16} x + \frac{81}{64} ln (4 x + 9) + c_{1}

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What is the solution for f^'(x)=(x^2)/((9+4x)) ?
The solution for f^'(x)=(x^2)/((9+4x)) is f(x)=(x^2)/8-9/16 x+81/64 ln(4x+9)+c_{1}