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I've been struggling with this one for a while, but couldn't do it. $$\int{\frac{1}{x^4\arctan(x)+x^3+x^2\arctan(x)+x}}dx$$ What I tried was factoring like so: $$I = \int{\frac{1}{(x^2+1)(x+x^2\arctan(x))}}dx$$then substitute $u=\arctan(x)$, which leads to: $$I=\int{\frac{1}{\tan(u)+u\cdot\tan^2(u)}}du$$which I've tried to solve using trigonometric identities, or even substituting $t = \ln(\tan(u))$, without success.

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Your method so far is good, and is what I would have done too. The trick now is to write: $$\int \frac{1}{\tan(u)+u\cdot \tan^2(u)}~du=\int \frac{\cot^2(u)}{u+\cot(u)}~du=\int \frac{\csc^2(u)-1}{u+\cot(u)}~du$$ Where we have used the well-known identity $\cot^2 (\theta) \equiv \csc^2 (\theta)-1$. Now substitute $s=u+\cot(u)$ and thus $ds=1-\csc^2(u)~du$, and you will obtain: $$\int \frac{\csc^2(u)-1}{u+\cot(u)}~du=-\int \frac{1}{s}~ds$$ Which is easy to evaluate.

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    $\begingroup$ Your u-sub is not right. It should be $ds=(1-cot^2(u))du$, and that changes everything $\endgroup$ – imranfat Oct 12 '17 at 17:16
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    $\begingroup$ @imranfat Are you sure? $s=u+\cot(u)$ gives $\frac{ds}{du}=1-\csc^2(u)=-\cot^2(u)$ after using the identity $\cot^2(u)\equiv \csc^2(u)-1$. $\endgroup$ – projectilemotion Oct 12 '17 at 17:25
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    $\begingroup$ Ah, those identities.... $\endgroup$ – imranfat Oct 12 '17 at 17:26
  • $\begingroup$ Holy smokes I would have never seen that. $\endgroup$ – Randall Oct 12 '17 at 17:30

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