Answer
an explicit formula for $f(x)$ is
$$
f(x) =\frac{1+x^{2}}{x^{2}}(x \cos x+\sin x)
$$
Work Step by Step
$$
\int_{0}^{x} f(t) d t=x \sin x+\int_{0}^{x} \frac{f(t)}{1+t^{2}} d t
$$
by differentiation we get:
$$
\begin{aligned}
\frac{d}{dx} \left[ \int_{0}^{x} f(t) d t \right] &=\frac{d}{dx} \left[x \sin x+\int_{0}^{x} \frac{f(t)}{1+t^{2}} d t \right]\\
f(x)&=x \cos x+\sin x+\frac{f(x)}{1+x^{2}}\\
f(x)\left(1-\frac{1}{1+x^{2}}\right) &=x \cos x+\sin x\\
f(x)\left(\frac{x^{2}}{1+x^{2}}\right) &=x \cos x+\sin x \\
f(x) &=\frac{1+x^{2}}{x^{2}}(x \cos x+\sin x)
\end{aligned}
$$
Thus an explicit formula for $f(x)$ is
$$
f(x) =\frac{1+x^{2}}{x^{2}}(x \cos x+\sin x)
$$
