## appleduardo 3 years ago whats the integral of x arctan x dx??

1. ksaimouli

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2. ksaimouli

u need to remember this $\frac{ 1 }{ \sqrt{x^2+1} }$+c

3. anonymous

How are you doing on this one? Do you still want help?

4. anonymous

$\int x\arctan x\;dx$ 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. anonymous

$\large \frac{x^2}{x^2+1}=\frac{(x^2+1)-1}{x^1+1}=1-\frac{1}{x^2+1}$

6. appleduardo

what trig substitution do I have to apply?

7. anonymous

|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. anonymous

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.