Assuming you mean
[tex]f'(x)=\sqrt x(3-10x},\,f(1)=9[/tex]
Integrate both sides to get
[tex]\displaystyle\int f'(x)\,\mathrm dx=f(x)=\int\sqrt x(3-10x)\,\mathrm dx[/tex]
You have
[tex]f(x)=\displaystyle\int\sqrt x(3-10x)\,\mathrm dx=\int(3x^{1/2}-10x^{3/2})\,\mathrm dx=2x^{3/2}-4x^{5/2}+C[/tex]
Since [tex]f(1)=9[/tex], you have
[tex]9=2\times1^{3/2}-4\times1^{5/2}+C\implies C=11[/tex]
so the particular solution is
[tex]f(x)=2x^{3/2}-4x^{5/2}+11=2x^{3/2}(1-4x)+11[/tex]
