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Evaluate this integral? $$\int_{0}^{1}\left(\frac{\arctan(x)}{1+x\arctan(x)}\right)^2\mathrm dx$$

$y=\arctan(x)$

$\frac{dy}{dx}=\frac{1}{1+\tan^2(y)}$

$$\int_{0}^{\pi/4}\frac{y^2(1+\tan^2(y))}{(1+y\tan(y))^2}\mathrm dy$$

This looked transformation is far harder than the original one.

I checked on Wolfram integral, it returns a nice closed form of $\frac{4-\pi}{4+\pi}$ apparently. Is this result correct?

  • $$\int{{{\left( \frac{{{\tan }^{-1}}\left( x \right)}{1+x{{\tan }^{-1}}\left( x \right)} \right)}^{2}}dx}=\frac{x-{{\tan }^{-1}}\left( x \right)}{x{{\tan }^{-1}}\left( x \right)+1}+C$$ – logo Jun 04 '19 at 04:19
  • Computer algebra gives: $\frac{4-\pi }{4+\pi }$. – David G. Stork Jun 04 '19 at 04:34

1 Answers1

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Hint

Multiply numerator and denominator by $\cos^2y$ to find

$$\int\dfrac{y^2}{(\cos y+y\sin y)^2}dy$$

Now as $\dfrac{d(\cos y+y\sin y)}{dy}=y\cos y$

write $\dfrac{y^2}{(...)^2}=\dfrac{y\cos y}{(...)^2}\cdot y\sec y$

Now integrate by part

lab bhattacharjee
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