What is the solution for f^'(x)=sqrt(x+4) ?

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

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f^'(x)=sqrt(x+4)

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

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

Solution steps

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

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

then

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

f (x) = \int x + 4 d x

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

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

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