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I want to calculate the following integral $$\int_{0}^{1}\left(\frac{\arctan x}{1+(x+\frac{1}{x})\arctan x}\right)^2 \, dx$$

But I have no way to do it, can someone help me, thank you.

FofX
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    Previous episode: https://math.stackexchange.com/questions/2606178/how-to-compute-int-01-fracx-arctan-x1x-frac1x-arctan-xdx – Robert Z Jan 16 '18 at 14:24
  • @Robert Z - Indeed. Though despite appearances, the two integrals are actually quite different. – omegadot Jan 17 '18 at 23:50
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    Can someone tell me why this was downvoted? It was negative until I upvoted it. I had a great time trying and failing to solve this. – BenB Jan 18 '18 at 01:47
  • @BenB - I thought the same so gave it an upvote from me. – omegadot Jan 18 '18 at 07:30

3 Answers3

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Here is a way to arrive at the answer, and it is by no means obvious.

In finding the indefinite integral the reverse quotient rule will be used. Recall that if $u$ and $v$ are differentiable functions, from the quotient rule $$\left (\frac{u}{v} \right )' = \frac{u' v - v' u}{v^2},$$ it is immediate that $$\int \frac{u' v - v' u}{v^2} \, dx = \int \left (\frac{u}{v} \right )' \, dx = \frac{u}{v} + C. \tag1$$

Now the indefinite integral $$I = \int \left (\frac{\arctan (x)}{1 + \left (x + \frac{1}{x} \right ) \arctan (x)} \right )^2 \, dx,$$ can be rewritten as $$I = \int \frac{x^2 \arctan^2 (x)}{\left (x + (x^2 + 1) \arctan (x) \right )^2} \, dx.$$

Let $v = x + (x^2 + 1) \arctan (x)$. Then $v' = 2 + 2x \arctan (x)$. Now for the hard bit. We need to find a function $u(x)$ such that $$u' v - v' u = u'[x + (x^2 + 1) \arctan (x)] - u [2 - 2x \arctan (x)] = x^2 \arctan^2 (x).$$ If $$u = \frac{-x^2 + (x^2 + 1) \arctan^2 (x)}{2},$$ then $$u' = x + x \arctan^2 (x) + \arctan (x),$$ and we find, miraculously, that $$u' v - v' u = x^2 \arctan^2 (x).$$

Our indefinite integral can now be readily found as it can be rewritten in the form given by (1). The result is: $$I = \int \left (\frac{-x^2 + (1 + x^2) \arctan^2 (x)}{2[x + (x^2 + 1) \arctan (x)]} \right )' \, dx = \frac{-x^2 + (1 + x^2) \arctan^2 (x)}{2[x + (x^2 + 1) \arctan (x)]} + C.$$

So for the definite integral on the interval $x \in [0,1]$ we have $$\int_0^1 \left (\frac{\arctan (x)}{1 + \left (x + \frac{1}{x} \right ) \arctan (x)} \right )^2 \, dx = \frac{\pi^2 - 8}{8(2 + \pi)}.$$

omegadot
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  • Friend, thank you for this. I spent half a day on this to no avail. – BenB Jan 18 '18 at 05:06
  • @omegadot Well done! (+1) – Robert Z Jan 18 '18 at 07:29
  • @omegadot Thanks for your answer,but I don't understand how that u came out of it? – FofX Jan 18 '18 at 08:58
  • @FofX - Conjuring up a function $u(x)$ such that in our case $u' v - v' u = x^2 \arctan^2 (x)$ was by no means easy and is what usually places very limits on the effectiveness of using the reverse quotient rule as a method of integration. Sometimes the method may be easy to apply where the function $u(x)$ can be more or less immediately found by inspection but in this particular case it involved a lot of trail and error. Perhaps there is an alternative method for finding this particular integral that is more "obvious" but I have so far been unable to find one. – omegadot Jan 18 '18 at 23:21
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In fact, there exists a closed-form expression for the indefinite integral (credit Mathematica, not me, for this): $${-x^2+(1+x^2)(\arctan(x))^2\over 2(x+(1+x^2)\arctan(x))}\,,$$ as can be verified by differentiation. Integral over $x\in [0,1]$ is equal to ${\pi^2-8\over 8(2+\pi)}\approx0.0454529$.

user15994
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This is about as far as I got. Let me know if anyone can finish this analytically:

$$ \left( \frac{\arctan x}{1 + (x + \frac{1}{x}) \arctan x} \right)^2 = \\ \left( \frac{\frac{x}{1 + x^2} \cdot \arctan x}{x \cdot \frac{1}{1 + x^2} + \arctan x} \right)^2 = \\ \left( \frac{x \arctan x \cdot (\arctan x)'}{(x \arctan x)'} \right)^2 = \\ \left( \frac{(\arctan x)'}{[\ln(x \arctan x)]'} \right)^2 $$

BenB
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