Power series representation of $(1+x^2)\arctan x$ so i've been looking for a cleaner way of solving this question, finding a Power series representation of the following:
$$
\left(1+x^2\right)\arctan\left(x\right) 
$$
knowing the power series of arctan i got:
$$
\sum_{n=0}^{\infty \:}\frac{\left(-1\right)^n\cdot x^{2n+1}}{2n+1}\:+\:\sum_{n=0}^{\infty}\frac{\left(-1\right)^n\cdot \:x^{2n+3}}{2n+1}
$$
which after opening up the sum expressions amounts to :
$$
x+\:\:\sum_{n=1}^{\infty}\frac{\left(-1\right)^n\cdot \:2\cdot x^{2n+1}}{\left(2n\right)^2-1}
$$
I am sure there is a better way of doing this, maybe differentiation or integration? I've tried but i can't figure it out.
Any help would be greatly appreciated :]
 A: Yes, there is indeed a solution by differentiation + integration:
Differentiate your expression:
$$2x \arctan(x) +1$$
replace $\arctan(x)$ by its expansion, then integrate it term by term... taking into account the integration constant in such a way that there is an agreement in $x=0$ of the initial expression and your expansion.
A: If we have a polynomial of the form:
$$ P(x) = \sum a_i x^i$$
Another one of form:
$$ Q(x)  = \sum b_j x^j$$
$$ P(x) \cdot Q(x) = \sum a_i \cdot b_j x^{i+j}$$
Sub :
$$ i+j = u$$
Leads to:
$$ i= u-j$$
$$ P(x) \cdot Q(x)  = \sum_{u=0}^{\infty} \sum_{j=0}^u a_{u-j} b_j x^u$$
Or, r.h.s:
$$ \sum_{u=0}^{\infty} x^u \sum_{j=0}^u a_{u-j} b_j$$
Now call:
$$  \sum_{j=0}^u a_{u-j} b_j =  c_u$$
Or,
$$ \sum_{j=0}^u a_j b_{u-j} = c_u$$
For this particular case:
$$ a= \{ a_0 , a_1, a_2 ,0  ,0 , 0 ,0 ,0... \} =  \{ 1 , 0 , 1 , 0 , 0 , 0 ....\}$$
Hence:
$$  (1) b_{u} + (1)b_{u-2} = c_u$$
Now:
$$ u \to 2k+1$$
Or,
$$ b_{2k+1}  + b_{2k-1} = c_{2k+1}$$
For the $ \tan^{-1} $ series:
$$ b_{2k+1} = \frac{(-1)^k}{2k+1}$$
And,
$$ b_{2k-1} = \frac{ (-1)^{k-1} }{ 2k-1}$$
Hence:
$$ c_{2k+1} = \frac{(-1)^{k-1}}{2k-1} + \frac{ (-1)^k}{2k+1}$$
Hence our series is:
$$ \sum_{k=0}^{\infty} c_{2k+1} x^{2k+1}$$
DONE!
