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

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

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

SithsAndGigglesBest 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

sirm3dBest 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

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

jumping_mouseBest 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_mouseBest 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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