# Find $\int{\frac{1}{x^4\arctan(x)+x^3+x^2\arctan(x)+x}}dx$

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.

## 1 Answer

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.

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