# Power Series of Integral xarctan(3x)

1. Oct 25, 2009

### EstimatedEyes

1. The problem statement, all variables and given/known data

Evaluate the integral of xarctan(3x) from 0 to 0.1 by expressing the integral in terms of a power series.

2. Relevant equations

3. The attempt at a solution

I differentiated xarctan(3x) until I got two functions that I could turn into power series (arctan(3x) and x(1/(1+9x^2)).

After I found the power series of these two functions I integrated them as many times as was necessary to arrive back at the integral of the original function and got (both as sums from one to infinity):
[(-1)^(n-1)*9^(n-1)*x^(2n+1)]/[(2n+1)(2n)(2n-1)] + [(-1)^(n-1)*9^(n-1)*x^(2n)]/[(2n)(2n-1)]

When I evaluated it from 0 to 0.1, I ended up with 0.0050927 which is not the correct answer, but I don't see where I would have done it wrong. Any help would be appreciated.

2. Oct 25, 2009

### zcd

A faster way to find the power series would be knowing $$\arctan{u}=\sum_{n=0}^{\infty}(-1)^{n}\frac{u^{2n+1}}{2n+1}$$ convergent for $$|x|\leq{1}$$. $$x\arctan{3x}=\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{2n+1}}{2n+1}x^{2(n+1)}$$. I haven't checked your work but maybe you made a mistake while differentiating.

On a separate note, $$\int \frac{x}{1+9x^{2}}\,dx=\frac{1}{18}\ln{|1+9x^{2}|}+C$$, not the arctan you were looking for.