# Find the constant value of the difference

I know that the following two functions have the same derivative: ##\arctan (x-1)## and ##2 \arctan (x-1 + \sqrt{(x-1)^2+1})##. Out of curiosity, how can I find the constant value at which they differ? I tried to add ##\pi / 2## to arctan(x-1) but I'm not sure if that works or not...

member 587159
Don't know if this works, but if they are equal as functions, they are equal in every point on the domain. So try to find a good x to evaluate the functions in.

Don't know if this works, but if they are equal as functions, they are equal in every point on the domain. So try to find a good x to evaluate the functions in.
So in that case does ##\arctan (x-1) + \pi / 2 =2 \arctan (x-1 + \sqrt{(x-1)^2+1})##? I'm not sure how to check it.

member 587159
Let me be a little more specific. You know ##f'(x) = g'(x)##

Hence there exists a constant c such that ##f(x) = c + g(x)##

Now, you can start to find that constant by plugging in values of x.

In your example, put for example ##x = 1##.

Then ##0 = \arctan(0) = c + 2\arctan(1)##

Hence, ##c = -2 \arctan(1) = -2 \pi/4 = - \pi/2##

and indeed, ##\pi/2## works if you add it on the left.

Mr Davis 97
The input $x-1$ is an unnecessary complication for the actual result, and we could also state that the formula

$$D_x \arctan\big(x + \sqrt{1 + x^2}\big) = \frac{1}{2}\frac{1}{1 + x^2}$$

is true.