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

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

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

SithsAndGiggles
 one year ago
Best 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.

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

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

jumping_mouse
 one year ago
Best ResponseYou've already chosen the best response.1dw: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).

jumping_mouse
 one year ago
Best ResponseYou've already chosen the best response.1So 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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