Series expansion for $\arctan(1-x)$ Series expansion for $\arctan(1-x)$
I try to expand this function into its Taylor series by means of differentiating it and then integrating it terms by terms but I fail to obtain the correct result.
The derivative of $\arctan(1-x)=-\dfrac{1}{x^2-2x+2}$. By using long division, I can obtain the series expansion for $-\dfrac{1}{x^2-2x+2}$ and it is 
$-\dfrac{1}{2}-\dfrac{x}{2}-\dfrac{x^2}{4}+\dfrac{x^4}{8}+\dfrac{x^5}{8}+\dfrac{x^6}{16}...$
I integrate this series terms by terms and I obtain:
$0-\dfrac{x}{2}-\dfrac{x^2}{4}-\dfrac{x^3}{12}+\dfrac{x^5}{40}+\dfrac{x^6}{48}-\dfrac{x^7}{122}...$
The series according to Wolfram is: $\dfrac{\pi}{4}-\dfrac{x}{2}-\dfrac{x^2}{4}-\dfrac{x^3}{12}+\dfrac{x^5}{40}+\dfrac{x^6}{48}-\dfrac{x^7}{122}...$.
I notice that if I use the indefinite integral, there is going to be an arbitrary constant left after the integrating process. How do I obtain this $\dfrac{\pi}{4}$? Without it, is my series wrong?
Is there any other methods to expand this function into Maclaurin series without directly employing the Maclaurin formula

When you integrate it, you get a $+C$. To get it's value, you can substitute in $x=0$ to get that $C=\arctan(1)$.
