Power series for $\ln(1+x^2)$ In the problem I am asked to use a power series representation of  $\ln(1+x)$ to approximate the integral from $0$ to $0.5$ of $\ln(1+x^2)$ to within 4 decimal places.
So far I have found a series for $\ln(1+x^2)$ by manipulating the known series 1/(1-r)
$\ln (1+x)=x−x^2/2+(2x^3)/3−(3x^4)/4+(4x^5)/5 \cdots$
Substituting  x  for $ x^2$  yields:
$\ln (1+(x^2 ))=(x^2 )−(x^2 )^2/2+(2(x^2 )^3)/3−(3(x^2 )^4)/4+(4(x^2 )^5)/5−…=x^2−x^4/2+(2x^6)/3−(3x^8)/4+(4x^{10})/5 - \cdots$
In order to determine accuracy I have to figure out how many terms to take and to figure this out I must be able to define a general term in the series. I have tried for several hours to determine the general terms, and the closest I have come to success is a general term which describes the entire series that I can imagine but not the first term. Something like:$\sum_{n=1}^∞ (−1)^n∙2(n^2+n) x^{2n+3})/(n+1)(2n+3)$
Can anybody tell if/where there is an error in my arithmetic/approach that is causing problems, and send me in the right direction? Or, if anybody knows of any standard procedure to obtain the general term from this point, this would be helpful as well.

Recall that $\frac{d}{dx}\ln(1+x^2)=\frac{2x}{1+x^2}=2x\cdot\sum_{i=0}^{\infty}(-1)^ix^{2i}=2\cdot \sum_{i=0}^{\infty}(-1)^ix^{2i+1}$, so $\ln(1+x^2)=\int 2\cdot \sum_{i=0}^{\infty}(-1)^ix^{2i+1}dx=2\cdot \sum_{i=0}^{\infty}(-1)^i\frac{x^{2i+2}}{2i+2}$.
