<s>[INST] Show that $\int_{0}^x \dfrac{1}{1+t^2}dt+\int_{0}^{1/x} \dfrac{1}{1+t^2}dt$ does not depend on $x$ 
Show that $\displaystyle \int_{0}^x \dfrac{1}{1+t^2}dt+\int_{0}^{1/x} \dfrac{1}{1+t^2}dt$ does not depend on $x$.

Attempt:
We have that $\displaystyle \int_{0}^x \dfrac{1}{1+t^2}dt+\int_{0}^{1/x} \dfrac{1}{1+t^2}dt = \tan^{-1}{x}+\tan^{-1}{\dfrac{1}{x}}$, which does depend $x$, so I am confused.
 [/INST] So many answers here to choose from.  Here is a slightly different approach.
Here you are:
$\tan^{-1} x + \tan^{-1} \frac{1}{x} + C\\
x = \tan y, -\frac{pi}{2}<y<\frac{pi}{2}\\
\tan^{-1} (\tan y) + \tan^{-1} (\frac{1}{\tan y}) + C\\
y + \tan^{-1} (\cot y) + C\\
y + \tan^{-1} (\tan (\frac{\pi}{2}- y) + C\\
\frac{\pi}{2} + C$
Actually,if $x<0,$ and hence $y<0$ then $\frac{\pi}{2} - y > \frac{\pi}{2}$ and $\tan^{-1} (\tan (\frac{\pi}{2} - y)) = -\frac{\pi}{2} - y$
hence  $\tan^{-1} x + \tan^{-1} \frac{1}{x} =-\frac{\pi}{2},\frac{\pi}{2}$ depending on the sign of $x$.
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