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appleduardo

  • 3 years ago

whats the integral of x arctan x dx??

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  1. ksaimouli
    • 3 years ago
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    |dw:1363212783612:dw|

  2. ksaimouli
    • 3 years ago
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    u need to remember this \[\frac{ 1 }{ \sqrt{x^2+1} }\]+c

  3. jumping_mouse
    • 3 years ago
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    How are you doing on this one? Do you still want help?

  4. SithsAndGiggles
    • 3 years ago
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    \[\int x\arctan x\;dx\] 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.

  5. sirm3d
    • 3 years ago
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    \[\large \frac{x^2}{x^2+1}=\frac{(x^2+1)-1}{x^1+1}=1-\frac{1}{x^2+1}\]

  6. appleduardo
    • 3 years ago
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    what trig substitution do I have to apply?

  7. jumping_mouse
    • 3 years ago
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    |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).

  8. jumping_mouse
    • 3 years ago
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    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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