Answer
$x^3-\dfrac{1}{3}x^7+\dfrac{1}{5}x^{11}-....$
or,
$\Sigma_{n=1}^\infty (-1)^{(n+1)}\dfrac{x^{(4n-1)}}{(2n-1)}$
Work Step by Step
Consider the Taylor Series for $\tan^{-1} x$ as follows:
$\tan^{-1} x=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-....$
Now, $ x\tan^{-1} x^2=x[x^2-\dfrac{(x^6)}{3}+\dfrac{(x^{10})}{5}-....]=x^3-\dfrac{1}{3}x^7+\dfrac{1}{5}x^{11}-....$
This can be written as: $\Sigma_{n=1}^\infty (-1)^{(n+1)}\dfrac{x^{(4n-1)}}{(2n-1)}$
