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- Thread starter Mr Davis 97
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- #1

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- #2

member 587159

- #3

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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.

- #4

member 587159

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.

- #5

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[tex]

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

[/tex]

is true.

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