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ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
dw:1363212783612:dw
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
u need to remember this \[\frac{ 1 }{ \sqrt{x^2+1} }\]+c
 one year ago

jumping_mouse Group TitleBest ResponseYou've already chosen the best response.1
How are you doing on this one? Do you still want help?
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
\[\int x\arctan x\;dx\] 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.
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
\[\large \frac{x^2}{x^2+1}=\frac{(x^2+1)1}{x^1+1}=1\frac{1}{x^2+1}\]
 one year ago

appleduardo Group TitleBest ResponseYou've already chosen the best response.0
what trig substitution do I have to apply?
 one year ago

jumping_mouse Group TitleBest ResponseYou've already chosen the best response.1
dw:1363282174743:dwImagine 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).
 one year ago

jumping_mouse Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago
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