Prove that $\int_{x}^{1} \frac{1}{1+ t^2} = \int_{1}^{\frac{1}{x}} \frac{1}{1+ t^2}$ for $x \gt 0$ Prove that $$\int_{x}^{1} \frac{1}{1+ t^2} = \int_{1}^{\frac{1}{x}} \frac{1}{1+ t^2} $$ for $x \gt 0$
Basically what i did is to form the function $$f(x)=\int_{x}^{1} \frac{1}{1+ t^2} - \int_{1}^{\frac{1}{x}} \frac{1}{1+ t^2}$$ and the I differentiated using the FTC to which i have
$$(\frac{1}{1+(\frac{1}{x})^2})\frac{1}{x^2} - \frac{1}{1+x^2} = 0$$ doing some algebra.
So then I know that the function that I made has to be a constant $\forall x \in \Bbb R$ but the problem is equality between the integrals is valid only for $x \gt 0$.
I know that at $0$ the function that  I made changes the value, but I don't know how to prove that it changes at $0$ and that for $x \gt 0$ the value is the same ($0$). Thanks in advance.
 A: Let $A = \Bbb{R}\setminus\{0\}$. Then, the function you defined is $f:A \to \Bbb{R}$. This function has the property that for every $x\in A$, $f'(x) = 0$. However, from this, you cannot conclude that $f$ is constant. All you can conclude is that $f$ is constant on each connected component of $A$ (look carefully at the proof of this theorem and try to find where the connectedness assumption comes into play). In this case, there are two connected components, namely $A_- = (-\infty, 0)$ and $A_+= (0,\infty)$. So, all you know is that $f|_{A_+}$ and $f|_{A_-}$ are constant. To evaluate what is the constant value of $f|_{A_+}$, simply evaluate at $x=1$, and you'll see that $f|_{A_+} = 0$.
However, there is no reason at all to expect that $f|_{A_-} = 0$. But since $f|_{A_-}$ is constant and $-1\in A_-$, we can evaluate this constant as follows: for every $x\in A_-$, we have
\begin{align}
f(x) &= f(-1) \\
&=\int_{-1}^1\dfrac{dt}{1+t^2} - \int_1^{-1}\dfrac{dt}{1+t^2} \\
&= 2 \int_{-1}^1 \dfrac{dt}{1+t^2} \\
&= 4\int_0^1 \dfrac{dt}{1+t^2} \\
&= 4 \arctan(t)\bigg|_0^{1} \\
&= \pi
\end{align}
Thus, $f$ is not a constant function. It is in fact given by
\begin{align}
f(x) &=
\begin{cases}
0 & \text{if $x\in A_+$} \\
\pi & \text{if $x \in A_-$}
\end{cases}
\end{align}
A: The function $ g: t\mapsto \frac{1}{1+t^2} $ is continuous at $ \Bbb R$ .So, the function $ x\mapsto \int_1^xg $ is differentiable at $ \Bbb R$ and your function $ f $ is differentiable at $ (0,+\infty) $ with
$$(\forall x>0)\;\; f'(x)=0$$
thus $ f $ is constante at $ (0,+\infty)$ and
$$(\forall x>0)\;\; f(x)=C=f(1)=$$
$$=\int_1^1g-\int_1^1g=0$$
Remark
You can prove your result using the substitution $$t=\frac 1u$$
$$\int_x^1\frac{dt}{1+t^2}=$$
$$\int_{\frac 1x}^1\frac{-\frac{du}{u^2}}{1+\frac{1}{u^2}}=$$
$$\int_1^{\frac 1x}\frac{du}{1+u^2}$$
A: Letting $s = 1/t$,
then $t = 1/s$ so
$dt = -ds/s^2$
so that
$\begin{array}\\
\int_{x}^{1} \dfrac{dt}{1+ t^2}
&=\int_{1/x}^{1} \dfrac{-ds/s^2}{1+ 1/s^2}\\
&=\int^{1/x}_{1} \dfrac{ds}{1+ s^2}\\
\end{array}
$
which is what you want.
Note that,
more generally,
$\begin{array}\\
\int_{a}^{b} \dfrac{dt}{1+ t^2}
&=\int_{1/a}^{1/b} \dfrac{-ds/s^2}{1+ 1/s^2}\\
&=\int^{1/a}_{1/b} \dfrac{ds}{1+ s^2}\\
\end{array}
$
A: We have (for a suitable function $f$):
$$\left(\int_x^1f(t)\,dt\right)'=\bigl(F(1)-F(x)\bigr)'=-f(x)$$
and
$$\left(\int_{1/x}^1f(t)\,dt\right)=\bigl(F(1/x)-F(1)\bigr)'=-f(1/x)\cdot\left(\frac{1}{x^2}\right).$$
Now let $f(x)=1/(1+x^2)$ and notice that$\int_1^1f(t)\,dt=0$ .
