12.25 For the function $f(x)=2 \sin 5 x+\sqrt{x}+\frac{3}{5} \quad$ find the antiderivative $F(x)$ under the condition that the graphs of the functions $f(x)$ and $F(x)$ intersect at a point lying on the $O y$ axis.
12.25 Since for the function $\sin x$ one of the antiderivatives is $-\cos x$, then, according to the rules of point $3^{0}$, the antiderivative of the function $2 \sin 5 x$ is $-\frac{2}{5} \cos 5 x$. Further, the antiderivative of the function $\sqrt{x}+\frac{3}{5}$ is $\frac{2}{3} x \sqrt{x}+\frac{3}{5} x$. Then for $f(x)$ an antiderivative function is $F(x)=-\frac{2}{5} \cos 5 x+\frac{2}{3} x \sqrt{x}+\frac{3}{5} x+C \quad$ for any value of the constant $C$. It is necessary to find such a value of $C$ for which the graphs of the functions $F(x)$ and $f(x)$ intersect at a point lying on the $O y$ axis. This means that at $x=0$ the equality
$F(0)=f(0) . \quad$ But $\quad F(0)=-\frac{2}{5}+C, \quad$ and $\quad f(0)=\frac{3}{5};$ therefore, $-\frac{2}{5}+C=\frac{3}{5}$, from which $C=1$.

Answer: $F(x)=-\frac{2}{5} \cos 5 x+\frac{2}{3} x \sqrt{x}+\frac{3}{5} x+1$.