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whats the integral of x arctan x dx??

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u need to remember this \[\frac{ 1 }{ \sqrt{x^2+1} }\]+c
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Other answers:

\[\int x\arctan x\;dx\] Integrate by parts, letting \[\begin{matrix}u=\arctan x&dv=x\;dx\\ du=\frac{1}{x^2+1}dx&v=\frac{1}{2}x^2\end{matrix}\] \[\frac{1}{2}x^2\arctan x-\frac{1}{2}\int\frac{x^2}{x^2+1}dx\] The new integral requires some trig substitution.
\[\large \frac{x^2}{x^2+1}=\frac{(x^2+1)-1}{x^1+1}=1-\frac{1}{x^2+1}\]
what trig substitution do I have to apply?
|dw:1363282174743:dw|Imagine a theta such that tan(theta) = x. Then you can draw this triangle. THen you have these facts: tan(theta) = x, theta = arctan(x) dx = sec(x)^2 d(theta).
So then you can substitute for your old integral: \[\int\limits_{}^{}\theta*\tan(\theta)*\sec(\theta)^2d \theta\] Let me know if you need more help that that.

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