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

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

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

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

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

Solution steps

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

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

then

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

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

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

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

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