Power Series of $\ \frac x3 \ln(1+x^2).$ I have a question regarding the power series representation of the function $$\ \frac x3 \ln(1+x^2).$$ 
I got the answer $$\sum_{n=1}^\infty \frac{(-1)^{n+1}{x^{2n+1}}}{3n} $$
by using the Maclaurin series expansion for $\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}{x^{n}}}{n}$ and replacing $x$ with $x^2$ and multiplying the series with $\frac{x}{3}$. 
However, my teacher got the answer $$\frac 23\sum_{n=1}^\infty \frac{(-1)^{n+1}{x^{2n+3}}}{2n+1}. $$
My teacher mentioned differentiation and integration, as one of the first steps in his method was taking $\ln(1+x^2)$ and representing it as $\int \frac{2x dx}{1+x^2}.$ He then represented the original function as $\frac x3 \int \frac{2x dx}{1+x^2}.$ However, after this I could not follow the rest of the procedures. I am completely lost as to how my teacher obtained this answer, and any help would be appreciated in explaining how one may arrive to this conclusion.

So continuing from where your teacher left off, we have $\frac{2x}3 \int\frac{xdx}{1+x^2}$. We start with the series expansion for $\frac{1}{1+y}$:
$$ \frac1{1+y} = \sum_{n=0}^\infty (-1)^ny^n$$
We substitute $y = x^2$:
$$ \frac1{1+x^2} = \sum_{n=0}^\infty (-1)^nx^{2n}$$
We multiply by $x$:
$$ \frac x{1+x^2} = \sum_{n=0}^\infty (-1)^nx^{2n+1}$$
And now we integrate:
$$\int\frac{xdx}{1+x^2} = \sum_{n=0}^\infty \frac{(-1)^nx^{2n+2}}{2n+2} $$
Finally, we multiply by $\frac{2x}3$:
$$ \frac{2x}3 \int\frac{xdx}{1+x^2} = \frac23 \sum_{n=0}^\infty \frac{(-1)^nx^{2n+3}}{2n+2}$$
I imagine this was the intended methodology, but the answer I get doesn't match your teacher's. In fact, it matches yours by making the change of index $m=n+1$:
$$\frac{2x}3 \int\frac{xdx}{1+x^2} = \frac23 \sum_{m=1}^\infty \frac{(-1)^{m-1}x^{2m+1}}{2m}= \frac13 \sum_{m=1}^\infty \frac{(-1)^{m+1}x^{2m+1}}{m}$$
