What is the solution for f^'(x)=sqrt(x)(3+5x),f(1)=7 ?

The solution for f^'(x)=sqrt(x)(3+5x),f(1)=7 is f(x)=2x^{3/2}+2x^{5/2}+3

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f^'(x)=sqrt(x)(3+5x),f(1)=7

f^{^{'}} (x) = x (3 + 5 x), f (1) = 7

f (x) = 2 x^{\frac{3}{2}} + 2 x^{\frac{5}{2}} + 3

Solution steps

f^{'} (x) = x (3 + 5 x)

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

then

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

f (x) = \int x (3 + 5 x) d x

\int x (3 + 5 x) d x = 2 x^{\frac{3}{2}} + 2 x^{\frac{5}{2}} + c_{1}

f (x) = 2 x^{\frac{3}{2}} + 2 x^{\frac{5}{2}} + c_{1}

Apply initial conditions:

f (x) = 2 x^{\frac{3}{2}} + 2 x^{\frac{5}{2}} + 3

f (x) = 2 x^{\frac{3}{2}} + 2 x^{\frac{5}{2}} + 3

\int \frac{sin ( x )}{sin ( x ) + cos ( x )} d x

f, tan (f) (x) = 7 x^{4} + 4 \frac{x}{2} x^{3} + 1

\int (\frac{1}{x ^{3}} + \frac{7}{x ^{2}}) d x

f (x) = x^{2} + 5, (2, 9)

10 1 + x

What is the solution for f^'(x)=sqrt(x)(3+5x),f(1)=7 ?
The solution for f^'(x)=sqrt(x)(3+5x),f(1)=7 is f(x)=2x^{3/2}+2x^{5/2}+3