For the function $f(x)=\int_0^x \frac{dt}{1+t^2}$, answer the questions as follows.

Note : Please solve the problems without using directly the formula $\int \frac{1}{1+x^2}\ dx=\tan^{-1}x +C$ for Japanese High School students those who don't study arc sin x, arc cos x, arc tanx.

(1) Find $f(\sqrt{3})$

(2) Find $\int_0^{\sqrt{3}} xf(x)\ dx$

(3) Prove that for $x>0$. $f(x)+f\left(\frac{1}{x}\right)$ is constant, then find the value.
### Part (1): Find \( f(\sqrt{3}) \)

1. The function \( f(x) \) is defined as:
   \[
   f(x) = \int_0^x \frac{dt}{1+t^2}
   \]
2. We need to evaluate \( f(\sqrt{3}) \):
   \[
   f(\sqrt{3}) = \int_0^{\sqrt{3}} \frac{dt}{1+t^2}
   \]
3. Recognize that the integral \(\int \frac{1}{1+t^2} dt\) is the antiderivative of \(\frac{1}{1+t^2}\), which is \(\arctan(t)\):
   \[
   f(\sqrt{3}) = \left[ \arctan(t) \right]_0^{\sqrt{3}}
   \]
4. Evaluate the definite integral:
   \[
   f(\sqrt{3}) = \arctan(\sqrt{3}) - \arctan(0)
   \]
5. Knowing that \(\arctan(\sqrt{3}) = \frac{\pi}{3}\) and \(\arctan(0) = 0\):
   \[
   f(\sqrt{3}) = \frac{\pi}{3} - 0 = \frac{\pi}{3}
   \]

### Part (2): Find \(\int_0^{\sqrt{3}} xf(x) \, dx\)

1. We need to calculate:
   \[
   \int_0^{\sqrt{3}} x f(x) \, dx
   \]
2. Substitute \( f(x) = \int_0^x \frac{dt}{1+t^2} \):
   \[
   \int_0^{\sqrt{3}} x \left( \int_0^x \frac{dt}{1+t^2} \right) dx
   \]
3. Use the given solution approach:
   \[
   \int_0^{\sqrt{3}} \frac{x}{1 + x^2} \, dx
   \]
4. Multiply by 1 in the form \(\frac{2}{2}\):
   \[
   \frac{1}{2} \int_0^{\sqrt{3}} \frac{2x}{1 + x^2} \, dx
   \]
5. Use the substitution \( u = 1 + x^2 \), then \( du = 2x \, dx \):
   \[
   \frac{1}{2} \int_0^{\sqrt{3}} \frac{1}{u} \, du
   \]
6. Change the limits of integration: when \( x = 0 \), \( u = 1 \); when \( x = \sqrt{3} \), \( u = 4 \):
   \[
   \frac{1}{2} \int_1^4 \frac{1}{u} \, du
   \]
7. Integrate with respect to \( u \):
   \[
   \frac{1}{2} \left[ \ln u \right]_1^4 = \frac{1}{2} \left( \ln 4 - \ln 1 \right)
   \]
8. Simplify:
   \[
   \frac{1}{2} \ln 4 = \frac{\ln 4}{2}
   \]

### Part (3): Prove that for \( x > 0 \), \( f(x) + f\left(\frac{1}{x}\right) \) is constant, then find the value

1. Consider \( f(x) \) and \( f\left(\frac{1}{x}\right) \):
   \[
   f(x) = \int_0^x \frac{dt}{1+t^2}, \quad f\left(\frac{1}{x}\right) = \int_0^{\frac{1}{x}} \frac{dt}{1+t^2}
   \]
2. Add the two integrals:
   \[
   f(x) + f\left(\frac{1}{x}\right) = \int_0^x \frac{dt}{1+t^2} + \int_0^{\frac{1}{x}} \frac{dt}{1+t^2}
   \]
3. Change the variable in the second integral: let \( u = \frac{1}{t} \), then \( du = -\frac{1}{t^2} dt \):
   \[
   \int_0^{\frac{1}{x}} \frac{dt}{1+t^2} = \int_{\infty}^x \frac{-du}{1+\frac{1}{u^2}} = \int_x^{\infty} \frac{du}{1+u^2}
   \]
4. Combine the integrals:
   \[
   f(x) + f\left(\frac{1}{x}\right) = \int_0^x \frac{dt}{1+t^2} + \int_x^{\infty} \frac{du}{1+u^2}
   \]
5. Recognize that the sum of these integrals covers the entire range from 0 to \(\infty\):
   \[
   f(x) + f\left(\frac{1}{x}\right) = \int_0^{\infty} \frac{dt}{1+t^2}
   \]
6. The integral \(\int_0^{\infty} \frac{dt}{1+t^2}\) is known to be \(\frac{\pi}{2}\):
   \[
   f(x) + f\left(\frac{1}{x}\right) = \frac{\pi}{2}
   \]

The final answer is \( \boxed{ \frac{\pi}{2} } \)