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I need integrate $$\int \frac{\tan \sqrt{x}}{x\sqrt{x}} dx$$ I using $u = \sqrt{x}$, then $$2\int \frac{\tan u}{u^{2}} du$$

I try to use sustitution again, and for part, but a dont have answer.

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  • $\begingroup$ Is this a problem from a textbook? Any particular reason why this can be integrated in terms of elementary functions? $\endgroup$ – imranfat Oct 19 '16 at 2:53
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    $\begingroup$ There is no closed form: wolframalpha.com/input/?i=(tan+(sqrt+x))%2F(x%5E(3%2F2))+dx $\endgroup$ – MathematicsStudent1122 Oct 19 '16 at 2:53
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    $\begingroup$ @Wmmoreno. Are you sure (just checking) that it isn't an Arctan instead of just tan? Because then it is doable... $\endgroup$ – imranfat Oct 19 '16 at 2:56
  • $\begingroup$ @imranfat It is a integral examination of my chair of calculus $\endgroup$ – Wmmoreno Oct 19 '16 at 3:00
  • $\begingroup$ @Wmmoreno. Well, this one simply cannot be done....Of course, one can resort to series if needed, but if they want you to anti derive this one, it ain't gonna work in terms of standard functions... $\endgroup$ – imranfat Oct 19 '16 at 3:02
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As said in comments, the problem cannot be $$I=2\int \frac{ \tan (t)}{t^2}\,dt$$ which does not show analytical expression even using special function.

More than likely, they ask for $$I=2\int \frac{ \tan^{-1} (t)}{t^2}\,dt$$ which is easy to integrate by parts $$u=\tan^{-1} (t)\implies u'=\frac{dt}{t^2+1}$$ $$v'=\frac {dt} {t^2}\implies v=-\frac 1 t$$ All of that makes $$I=2\left(-\frac{\tan ^{-1}(t)}{t} +\int \frac{dt}{t \left(t^2+1\right)}\right)$$ For the last integral, partial fraction decomposition gives $$\frac{1}{t \left(t^2+1\right)}=\frac{1}{t}-\frac{t}{t^2+1}$$ which is easy to integrate.

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