# The antiderivative of 1/(x^4+1): Part 9

In the course of evaluating the antiderivative $\displaystyle \int \frac{1}{x^4 + 1} dx$,

I have stumbled across a very curious trigonometric identity: $\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi$ if $x < x_2$, $\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ if $x_2 < x < x_1$, $\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi$ if $x> x_1$,

where $x_1$ and $x_2$ are the unique values so that $\tan^{-1} ( x_1\sqrt{2} - 1 ) + \tan^{-1}( x_1 \sqrt{2} + 1) = \displaystyle \frac{\pi}{2}$, $\tan^{-1} ( x_2\sqrt{2} - 1 ) + \tan^{-1}( x_2 \sqrt{2} + 1) = -\displaystyle \frac{\pi}{2}$.

I will now show that $x_1 = 1$ and $x_2 = -1$. Indeed, it’s apparent that these have to be the two transition points because these are the points where $\displaystyle \frac{x \sqrt{2}}{1 - x^2}$ is undefined. However, it would be more convincing to show this directly.

To show that $x_1 = 1$, I need to show that $\tan^{-1} (\sqrt{2} - 1 ) + \tan^{-1}( \sqrt{2} + 1) = \displaystyle \frac{\pi}{2}$.

I could do this with a calculator… …but that would be cheating.

Instead, let $\alpha = \tan^{-1} (\sqrt{2} - 1 )$ and $\beta = \tan^{-1} (\sqrt{2} + 1 )$, so that $\tan \alpha = \sqrt{2} - 1$, $\tan \beta = \sqrt{2} + 1$.

Indeed, by SOHCAHTOA, the angles $\alpha$ and $\beta$ can be represented in the figure below: The two small right triangles make one large triangle, and I will show that the large triangle is also a right triangle. To do this, let’s find the lengths of the three sides of the large triangle. The length of the longest side is clearly $\sqrt{2} - 1 + \sqrt{2} + 1 = 2\sqrt{2}$. I will use the Pythagorean theorem to find the lengths of the other two sides. For the small right triangle containing $\alpha$, the missing side is $\sqrt{ \left(\sqrt{2} - 1 \right)^2 + 1^2} = \sqrt{2 - 2\sqrt{2} + 1 + 1} = \sqrt{4-2\sqrt{2}}$

Next, for the small right triangle containing $\beta$, the missing side is $\sqrt{ \left(\sqrt{2} + 1 \right)^2 + 1^2} = \sqrt{2 + 2\sqrt{2} + 1 + 1} = \sqrt{4+2\sqrt{2}}$

So let me redraw the figure, eliminating the altitude from the previous figure: Notice that the condition of the Pythagorean theorem is satisfied, since $\left( \sqrt{4-2\sqrt{2}} \right)^2 + \left( \sqrt{4+2\sqrt{2}} \right)^2 = 4 - 2\sqrt{2} + 4 + 2 \sqrt{2} = 8$,

or $\left( \sqrt{4-2\sqrt{2}} \right)^2 + \left( \sqrt{4+2\sqrt{2}} \right)^2 = \left( 2\sqrt{2} \right)^2$.

Therefore, by the converse of the Pythagorean theorem, the above figure must be a right triangle (albeit a right triangle with sides of unusual length), and so $\alpha + \beta = \pi/2$. In other words, $x_1 = 1$, as required.

To show that $x_2 = -1$, I will show that the function $f(x) = \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1)$ is an odd function using the fact that $\tan^{-1} x$ is also an odd function: $f(-x) = \tan^{-1} ( -x\sqrt{2} - 1 ) + \tan^{-1}( -x \sqrt{2} + 1)$ $= \tan^{-1} ( -[x\sqrt{2} + 1] ) + \tan^{-1}( -[x \sqrt{2} - 1])$ $= -\tan^{-1} ( x\sqrt{2} + 1 ) - \tan^{-1}( x \sqrt{2} - 1)$ $= - \left[ \tan^{-1} ( x\sqrt{2} + 1 ) + \tan^{-1}( x \sqrt{2} - 1) \right]$ $= -f(x)$.

Therefore, $f(-1) = -f(1) = -\displaystyle \frac{\pi}{2}$, and so $x_2 = -1$.

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