Answer
$2x^2-\dfrac{2^2x^3}{2}+\dfrac{2^3x^{4}}{3}-$
or, $\Sigma_{n=0}^\infty (-1)^n \dfrac{2^{n+1} x^{n+2}}{n+1}$
Work Step by Step
Since, we know that the Maclaurin Series for $ln(1+x)$ is defined as:
$ \ln (1+x)=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^{n+1}}{(n+1)!}=x-\dfrac{x^2}{2!}+\dfrac{x^3}{3!}-\dfrac{x^4}{4!}+...$
Now, $ x \cdot \ln (1+2x)=x[2x-\dfrac{(2x)^2}{2!}+\dfrac{(2x)^3}{3!}-\dfrac{(2x)^4}{4!}+...]$
Thus, $=2x^2-\dfrac{2^2x^3}{2}+\dfrac{2^3x^{4}}{3}-$
or, $\Sigma_{n=0}^\infty (-1)^n \dfrac{2^{n+1} x^{n+2}}{n+1}$
