Taylor series of $x \cdot \ln(10+x) $ with $x=9$ point of expansion Have been trying to solve this for quiet long, however still have no idea of how to.
$$x \cdot \ln(10+x) $$ 
with $x=9$ point of expansion.
Will appreciate any advice, thanks in advance!

$$f(x)=x \ln (19+x-9)= x \ln 19+ x\ln\left(1+\frac{(x-9)}{19}\right)$$
Let $(x-9)/19=z$, then
$$\implies f(x)= 19 z \ln 19+ 9 \ln 19 + 19z \ln(1+z)+9 \ln (1+z)$$
Now use $$\ln(1+z)=-z-z^2/2-z^3/2--...$$
to expand $f(x)$ in the powers of $z=(x-9)/19.$ and add the co-efficients of similar powers of $z$.WE get
$$f(x)=9\ln 19+ (9+19\ln 19) z+\frac{29}{2}z^2-\frac{13}{2} z^3+\frac{49}{12} z^4-..., z=(x-9)/19.$$
