Answer
$x^3-\dfrac{x^7}{3}+\dfrac{x^{11}}{5}-....$ or, $\Sigma_{n=1}^\infty (-1)^{n}\dfrac{x^{4n-1}}{2n-1}$
Work Step by Step
We know that the Taylor Series for $\tan^{-1} x$ is defined as:
$\tan^{-1} x=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-....$
Now, $ x\tan^{-1} x^2=x[x^2-\dfrac{x^6}{3}+\dfrac{x^{10}}{5}-....]$
or, $=x^3-\dfrac{x^7}{3}+\dfrac{x^{11}}{5}-....$ or, $\Sigma_{n=1}^\infty (-1)^{n}\dfrac{x^{4n-1}}{2n-1}$
