How to do this question that talks about dependency of x Let $x > 0$. Prove that the value of the following expression doesn't  depend on x
$$\int_{0}^{x} \frac{1}{1+t^2} dt + \int_{0}^{\frac{1}{x}} \frac{1}{1+t^2}dt$$
Attempt:
Left: f'(x) = $\frac{1}{1+x^2}$
Right: f'(x) = $\frac{1}{1+(\frac{1}{x})^2} -\frac{1}{x^2}$
$= \frac{1}{(1+\frac{1}{x^2})} - \frac{1}{x^2}$
$=\frac{x^2}{1+x^2} - \frac{1}{x^2}$
$=\frac{x^4 - x^2 - 1}{1+x^2}$
Yeah I don't know what I am doing, I tried to remove the integral but failed miserably

Using the $u$ sub $t' = t^{-1}$, we see  $dt'= -(t')^2dt$ that
\begin{align}
\int^{\frac{1}{x}}_0\frac{1}{1+t^2}\ dt = -\int^x_\infty \frac{1}{1+(t')^{-2}} \frac{dt'}{(t')^2}= \int^\infty_x \frac{1}{1+(t')^2}\ dt' = \frac{\pi}{2} - \int^x_0\frac{1}{1+(t')^2}\ dt'. 
\end{align}
But we see that
\begin{align}
\frac{\pi}{2} - \int^x_0\frac{1}{1+(t')^2}\ dt'= \int^x_0\frac{1}{1+(t')^2}\ dt'
\end{align}
provided
\begin{align}
\int^x_0\frac{1}{1+(t')^2}\ dt' = \frac{\pi}{4}.
\end{align}
Thus, your claim is incorrect since equality holds only when $x=1$. 
