<s>[INST] 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
 [/INST] To answer the question as-is after having been heavily edited, let for $x \gt 0\,$:
$$f(x) = \int_{0}^{x} \frac{1}{1+t^2} dt + \int_{0}^{\frac{1}{x}} \frac{1}{1+t^2}dt$$
Then, using the Leibniz integral rule:
$$
f'(x) = \frac{1}{1+x^2} \,-\, \frac{1}{x^2} \cdot \frac{1}{1+ \cfrac{1}{x^2}} = \frac{1}{1+x^2} \,-\, \frac{1}{1+x^2} \,=\, 0 
$$
Thus $f'(x)=0\,$, so $f(x)$ is a constant, and therefore does not depend on $x$.
P.S. Note to the OP:

Right: f'(x) = $\frac{1}{1+(\frac{1}{x})^2} -\frac{1}{x^2}$

This looks like you attempted to use Leibniz' rule, but misapplied it. The $-\,\frac{1}{x^2}$ derivative of the upper bound is multiplied with, not added to, the end value of the function being integrated.
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