1. indefinate integral/power series

integral (x - arctan^x)/(x^3) dx

need to evaluate the integral and what is the radius of convergence?

thanks if you can show this one.

2. Originally Posted by rcmango
integral (x - arctan^x)/(x^3) dx

need to evaluate the integral and what is the radius of convergence?

thanks if you can show this one.
Power series for arctan

$\displaystyle \tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + -\cdots$

so

$\displaystyle \frac{x -\tan^{-1} x}{x^3} = \frac{ \frac{x^3}{3} - \frac{x^5}{5} + \frac{x^7}{7} - + \cdots}{x^3}$

$\displaystyle = \frac{1}{3} - \frac{x^2}{5} + \frac{x^4}{7} - \frac{x^6}{9} \cdots$

so

$\displaystyle \int \frac{x -\tan^{-1} x}{x^3}\,dx = \int \frac{1}{3} - \frac{x^2}{5} + \frac{x^4}{7} - \frac{x^6}{9} \cdots \,dx= \frac{x}{3} - \frac{x^3}{3 \cdot 5} + \frac{x^5}{5 \cdot 7} - \frac{x^7}{7 \cdot 9} \cdots$

or

$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{2n-1}}{(2n-1)(2n+1)}$

Then use the ratio test on this series to find the radius of converge. For the interval, check the endpoints.

3. Hey thanks, your explanation was very nice, and it helped me understand what was really going on.