Answer
$$x+x^2+\dfrac{x^3}{2}-\dfrac{x^3}{3}+\dfrac{x^4}{6}-\dfrac{x^4}{3}+...$$
Work Step by Step
We have the Maclaurin series for $ \sin x$ and $e^x$ as follows
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$$
$$\tan^{-1} x=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}+...$$
Now, we have the Maclaurin series of $e^x \tan^{-1} x$ as:
$$e^x \tan^{-1} x=\\
=(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+)(x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}+)+...\\
=x+x^2+\dfrac{x^3}{2!}-(\dfrac{x^3}{3}+\dfrac{x^4}{3})+...\\=x+x^2+\dfrac{x^3}{2}-\dfrac{x^3}{3}+\dfrac{x^4}{6}-\dfrac{x^4}{3}+...$$
