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