What is the Maclaurin series for f(x)=ln(1+x2)f(x) = \ln(1+x^2)f(x)=ln(1+x2)?
∑n=1∞(−1)n−1x2nn\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n}}{n}∑n=1∞n(−1)n−1x2n
∑n=1∞(−1)nx2nn\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n}∑n=1∞n(−1)nx2n
∑n=0∞(−1)nx2nn!\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}∑n=0∞n!(−1)nx2n
∑n=1∞x2nn\sum_{n=1}^{\infty} \frac{x^{2n}}{n}∑n=1∞nx2n