Answer
$x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+...$
Work Step by Step
$y=e^{x}ln(1+x)$
$ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+....+\frac{x^{n}}{n}$
and $e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...$
$e^{x}ln(1+x)=(1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+..)(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+....+\frac{x^{n}}{n})$
$\approx x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+...$