Prove: $\forall x \gt 1, \arctan(\frac{1+x}{1-x}) = \arctan(x) - \frac{3\pi}{4}$ What I did is find the derivative of $\arctan\left(\frac{1+x}{1-x}\right)$: 
$$\frac{d\left[\arctan\left(\frac{1+x}{1-x}\right)\right]}{dx} = \frac{1}{1+x^2}$$
We can notice that that is the derivative of $arctan(x)$ as well.
So we can state the following:
$$\int\frac{d\left[\arctan\left(\frac{1+x}{1-x}\right)\right]}{dx}\cdot dx = \int\frac{d\arctan(x)}{dx}\cdot dx  $$
Now this is where I'm starting to have hesitations: We want to prove the equality $\forall x \gt 1 $, so what I think I should do is take those integrals from 1 to x. Indeed that works:
$$\int_1^x\frac{d\left[\arctan\left(\frac{1+x}{1-x}\right)\right]}{dx}\cdot dx = \int_1^x\frac{d\arctan(x)}{dx}\cdot dx   $$
If we simplify, we'll arrive to the desired equation. There are two problems with the above integral: at $x = 1, \arctan\left(\frac{1+x}{1-x}\right)$ is not defined. However, $$\lim_{x\to1^+}\arctan\left(\frac{1+x}{1-x}\right) = -\frac{\pi}{2}$$ 
Since we're not interested in x = 1, am I right in thinking that if the following equation is the correct formulation of the problem?
$$\lim_{h\to1^+}\left[\int_h^x\frac{d\left[\arctan\left(\frac{1+x}{1-x}\right)\right]}{dx}\cdot dx\right] = \lim_{h\to1^+}\left[\int_h^x\frac{d\arctan(x)}{dx}\cdot dx \right]$$ 

The other problem is that I don't think that the fact that two functions have equal area under their curve means they're necessarily equal to each other for all x. However if I prove that they are equal at a point $x_0$, and then prove that they are also equal at $x_0+\epsilon$, where $\epsilon\to0^+$, then I prove that they are equal for all x greater than $x_0$
Is my reasoning corret?
EDIT: I noticed that I made a rather big copying mistake. Everywhere where I wrote 
$$\int\arctan\left(\frac{1+x}{1-x}\right)\cdot dx $$
I actually meant:
$$\int\frac{d\left[\arctan\left(\frac{1+x}{1-x}\right)\right]}{dx}\cdot dx $$
Which does indeed change the meaning of my equations by A LOT 
 A: It's not necessary to use integrals. When two functions defined over an interval have the same derivative, they differ by a constant.
This is a consequence of the mean value theorem: if $f$ is a function having zero derivative over an interval, then, for $a,b$ in this interval,
$$
\frac{f(b)-f(a)}{b-a}=f'(c)=0
$$
so $f(a)=f(b)$ and $f$ is constant. So, if $f'(x)=g'(x)$ over an interval, the derivative of $F(x)=f(x)-g(x)$ is zero and so $f(x)-g(x)$ is constant.
You have correctly proved that the two functions have the same derivative. Therefore there exists a constant $k$ such that
$$
\arctan\frac{1+x}{1-x}=k+\arctan x
$$
for every $x>1$.
Now we want to determine $k$. With the limit at $\infty$ is the easiest way:
$$
\lim_{x\to\infty}\arctan\frac{1+x}{1-x}=\arctan(-1)=-\frac{\pi}{4}
$$
and
$$
\lim_{x\to\infty}(k+\arctan x)=k+\frac{\pi}{2}
$$
Therefore
$$
k+\frac{\pi}{2}=-\frac{\pi}{4}
$$
and so
$$
k=-\frac{3\pi}{4}
$$

If you consider the same for $x<1$, you have
$$
\arctan\frac{x+1}{x-1}=k_1+\arctan x
$$
for a constant $k_1$ possibly different from the above one. Indeed, if we set $x=0$, we get
$$
\arctan1=k_1+\arctan0
$$
so
$$
k_1=\arctan1=\frac{\pi}{4}
$$

(Edit: the question was edited to get rid of the error.)
Your argument with integrals is faulty. The two functions have the same derivative, but this does not mean their integrals are equal.
What you can say is that, for an arbitrary $x_0>1$,
Your argument with integrals can be made easier as follows: if $x_0>1$ is arbitrary,
$$
\arctan\frac{1+x}{1-x}-
\arctan\frac{1+x_0}{1-x_0}
=\int_{x_0}^x\frac{1}{1+t^2}\,dt
=\arctan x-\arctan x_0
$$
by applying the fundamental theorem of calculus.
Since this holds for every $x$ and every $x_0$, you can plug in a particular value of $x_0$ and get the required identity. Again, it's easier to do the limit for $x_0\to\infty$, so
$$
\arctan\frac{1+x}{1-x}+\frac{\pi}{4}=\arctan x-\frac{\pi}{2}
$$
A: I'd say demonstrating this type of identities is easier and  more essential if you just work with trigonometric definitions, instead of basing your reasoning on concepts such as integrals and the Mean Value Theorem. After all, provided that angles are in the first quadrant, we are 'simply' talking about ratios of right-angled triangles' sides. The rest can be deduced by symmetries.
Consider the following Figure.

$\triangle ABC$ is right-angled and such that $\overline{AC} = 1$ and $\overline{BC} = x>1$,
so that
\begin{equation} \alpha = \arctan x.\tag{1}\label{alpha}\end{equation}
Extend $AC$ to a segment $AD$, so that $\overline{CD} = x$. Let also $DH$ be the line perpendicular to $AB$. 
Note that
$$ \angle ADB = \frac{\pi}{4},$$
and therefore
\begin{equation}\beta = \frac{3}{4}\pi -\alpha.\tag{2}\label{rel}\end{equation} 
Since
$$\triangle ABC \sim \triangle ADH,$$
we obtain
$$\overline{DH} = x\overline{AH},$$
and, by Pythagorean Theorem on $\triangle ADH$,
\begin{eqnarray}
\overline{AH}^2 &=& \overline{AD}^2 - \overline{DH}^2=\\
 &=& (1+x)^2-\overline{AH}^2x^2,
\end{eqnarray} 
yielding
$$ \overline{AH} = \frac{x+1}{\sqrt{1+x^2}}$$
and
$$\overline{DH} = \frac{x(x+1)}{\sqrt{1+x^2}}.$$
Finally
\begin{eqnarray}
\overline{BH} &=& \overline{AB}-\overline{AH}=\\
&=& \frac{x(x-1)}{\sqrt{1+x^2}}.
\end{eqnarray}
We can conclude that
\begin{eqnarray}
\beta &=& \arctan\left(\frac{\overline{DH}}{\overline{BH}}\right)=\\
&=& \arctan\left(\frac{x+1}{x-1}\right).\tag{3}\label{beta}
\end{eqnarray}
Plugging \eqref{alpha} and \eqref{beta} into \eqref{rel}, and taking into account the odd symmetry of $\arctan(\cdot)$, leads to the desired result, i.e.
$$\arctan\left(\frac{1+x}{1-x}\right) = \arctan x - \frac{3}{4}\pi,$$
for $x>1$.
$\blacksquare$
A: Let $f$ be the function defined by
$f(x)=\arctan(\frac{1+x}{1-x})$.
for $|x|\neq1$.
We have
$f(\frac{1}{x})=-f(x)$ and
$f(-x)=\pm\frac{\pi}{2}-f(x)$.
Assume now $x>1$ and put
$x=\tan(t)$ with $t\in(\frac{\pi}{4},\frac{\pi}{2})$.
Thus
$f(x)=\arctan(\tan(\frac{\pi}{4}+t))$
$=\frac{\pi}{4}+t-\pi$
$=\arctan(x)-\frac{3\pi}{4}$.
For the other cases $(x<-1,-1<x<0,0<x<1)$  we use the properties of the function $f$ we started with.
