I want to evaluate the following integral:
$$\int_0^{\pi/2} \dfrac{dx}{(1+x^2)(1+\tan x)}$$
I tried integration by parts, $u = \arctan x$ that leads to $\int_0^{\arctan \pi/2} \dfrac{du}{1+\tan(\tan u)}$, but nothing seems to work...
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1I would use $x=arctan(u) $ – EDX May 25 '20 at 10:19
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2I'd bet good money that it's supposed to be $$\int_0^{\pi/2} \dfrac{dx}{(1+x^2)(1+\tan^{-1} x)}$$instead. – user790072 May 25 '20 at 10:21
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@user790072 I'll bet good money that you are right. Even wolfram alpha can't work it out (although I don't have pro computational time but still wolfram alpha can't find anything) – Mathsisfun May 25 '20 at 10:23