This integral is from integral

Find $$\int_0^\infty\tan\left(\frac x{\sqrt{x^3+x^2}}\right)\frac{\ln(1+\sqrt x)}xdx$$

I have get $$\int_0^\infty\tan\left(\frac x{\sqrt{x^3+x^2}}\right)\frac{\ln(1+\sqrt x)}xdx=\int_0^{\infty}\tan\left(\frac1{\sqrt{x+1}}\right)\frac{\ln(1+\sqrt x)}{x}dx$$ Let$$\dfrac{1}{\sqrt{x+1}}=t$$ that $$I=\int_{0}^{1}\dfrac{\tan{t}\ln{\left(1+\sqrt{\frac{1}{t^2}-1}\right)}}{t-t^3}dt$$ This integral is have closed form ?

  • $\begingroup$ I have to wonder whether this even exists... won't it oscillate too much as you approach zero? $\endgroup$ – user361424 Sep 4 '16 at 8:02
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    $\begingroup$ Why is this interesting ? (It's possible to construct infinite complicate integrals.) Something special like e.g. with math.stackexchange.com/questions/520657/… ? $\endgroup$ – user90369 Sep 4 '16 at 8:08
  • $\begingroup$ @user90369,the same as fell $\endgroup$ – communnites Sep 4 '16 at 8:40
  • $\begingroup$ what does WolframAlpha mean? $\endgroup$ – Dr. Sonnhard Graubner Sep 4 '16 at 8:54
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    $\begingroup$ Based on past experience with log integrals, I believe that the integral has a better chance of possessing a closed form if $tan$ were replaced with $tan^{-1}$. $\endgroup$ – nospoon Sep 4 '16 at 9:24

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