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Q:
How to get to the bottom of this problem?
I have the following problem:
\begin{align}
f(x) &= \frac{1}{x^2}\left(x^2-\frac{2}{\pi}+\frac{2}{\pi}\arctan\frac{1}{x}\right) \\
f(x) &= \frac{1}{x^2}\left(x^2-2+2\arctan\frac{1}{x}\right) \\
f(x) &= \frac{1}{x^2}\left(x^2-2+2\tan^{ -1}x\right) \\
f(x) &= \frac{1}{x^2}\left(x^2-2+2x\right) \\
f(x) &= \frac{1}{x^2}\left(x^2-2+2x\right) \\
f(x) &= 0
\end{align}
Is there any logical way to get the final answer in one step?
A:
Use the tangent half-angle substitution $\tan(x) = x$:
$
\begin{align}
f(x) &= \frac{1}{x^2}\left(x^2-\frac{2}{\pi}+\frac{2}{\pi}\arctan\frac{1}{x}\right) \\
&= \frac{1}{x^2}\left(x^2-\frac{2}{\pi}+\frac{2}{\pi}\frac{x}{1+x^2}\right) \\
&= \frac{1}{x^2}\left(x^2-\frac{2}{\pi}+\frac{2}{\pi}\frac{x(1+x^2)}{1+x^2+x^4}\right) \\
&= \frac{1}{x^2}\left(x^2-\frac{2}{\pi}+\frac{2}{\pi}\frac{x(1+x^2)}{1+x^2+x^4}\right) \\
&= \frac{1}{x^2}\left(x^2-\frac{2}{\pi}+\frac{2}{\pi}\frac{x(1+x^2)}{1+2x^2+x^4}\right) \\
&= \frac{1}{x^2}\left(x^2-\frac{2}{\pi}+\frac{2}{\pi}\frac{x(1+x^2)}{ 0b46394aab
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