# Power series representation

1. May 14, 2010

### nameVoid

$$\int \frac{x-arctanx}{x^3}dx$$
$$\frac{d}{dx}( x-arctanx ) = 1-\frac{1}{1+x^2}=\frac{x^2}{x^2+1}$$
$$= x^2 \sum_{n=0}^{\infty}(-1)^nx^{2n} = \sum_{n=0}^{\infty}(-1)^nx^{2n+2}$$
$$\int \sum_{n=0}^{\infty}(-1)^nx^{2n+2} dx = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+3}}{2n+3}+C$$
$$C=0?$$
$$\int \frac{\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+3}}{2n+3}}{x^3} dx$$
$$\int \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{2n+3}}dx$$
$$\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+3)(2n+1)}}$$
$$\sum_{n=1}^{\infty}(-1)^n\frac{x^{2n}}{(2n+2)(2n)}}+C$$
here i took d/dx converted to geomtric then integrated divided by x^3 then integrated again not sure how to deal with the ontants of integration in this case took c=0 on the first integral

2. May 14, 2010

### vela

Staff Emeritus
If you set x=0, you can show that C is, in fact, 0 as you assumed.
You're fine up until the last step. Note that in the next-to-the-last line, you have odd powers of x, and in the final line you have only even powers of x. The two expressions aren't equal to each other.

3. May 15, 2010

right..