I do not know how to format equations on here, so here is a picture of my attempt: 
Why is it that when trying to find the coefficients about the indicial root r=0 leads to undefined coefficients? The textbook claims both roots r = 1 and 0 have the same series solution? Unfortunately the textbook does not show the derivation.
On
To find the indicial equation, write the ODE as, $$ x^2 y'' + (x) y' + 0 = 0 $$ From this form, $b(x) = x$ and $c(x) = 0$, giving $$ r(r-1) + b_0 r + c_0 = r^2 - r + r = r(r - 1) = 0 $$ $$ r_1 = 1,\ r_2 = 0 $$ Using the larger root $r_1$, $$ y_1 = \sum_{m=0}^{\infty} a_{m}\ x^{m+1} = \sum_{m=1}^{\infty} a_{m-1}\ x^{m} $$ $$ xy_1'' = \sum_{m=1}^{\infty} a_{m}\ m(m+1) x^{m} $$ $$ a_{m} = \frac{-a_{m-1}}{(m)(m+1)} $$ $$ y_1 = x\left[ 1 - \frac{x}{2!} + \frac{x^2}{2 \cdot 3!} - \frac{x^3}{6 \cdot 4!} + \dots\right] = \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{(m-1)!\ m!}\ x^{m} $$ Since the two roots of the indicial equation differ by an integer, there exists a second root of the form, $$ y_2 = ky_1\ln(x) + \sum_{m=0}^{\infty}A_m x^m $$ $$ y_2'' = k\left[y_1'' \ln(x) - \frac{y_1}{x^2} + \frac{2y_1'}{x}\right] + \sum_{m=2}^{\infty} A_m\ m(m-1) x^{m-2} $$ Skipping the steps to find $y_1'$ and $y_1/x$, and consolidating terms, and noting that the term in square brackets is zero, $$ xy_2'' + y_2 = 0 = k\ln(x)\ [xy_1'' + y_1] - \frac{ky_1}{x} + 2ky_1' + 1 + \sum_{m=1}^{\infty}\{A_m + A_{m+1}\ m(m+1)\} x^m $$ $$ -(k+1) = k \left[\sum_{m=1}^{\infty}\frac{(-1)^m (2m + 1)} {m!\ (m+1)!}\ x^m\right] + \sum_{m=1}^{\infty}[A_m + A_{m+1}\ m(m+1)] x^m $$ Since the LHS contains all of the $x^0$ terms, $$ k = -1 $$ Using this value of $k$ in the RHS, and matching the coefficients of $x^1$ onwards, and setting $A_0 = 1$ by convention, $$ A_{m+1} = \left[ \frac{(-1)^{m} (2m+1)}{m!\ (m+1)!} - A_m \right] \ \frac{1}{m(m+1)} $$ This generates the series, $$ \sum_{m=1}^{\infty} A_m x^m = x - \frac{5}{4}x^2 + \frac{5}{18}x^3 + \dots $$ The second lineraly independent solution with a singular point at $x = 0$ is therefore, $$ y_2(x) = -\ln(x)\ y_1(x) + \sum_{m=1}^{\infty} A_m x^m$$ where the coefficients $\{A_1, A_2, A_3, \dots\}$ are found using the above recursive relation.
The alternative approach is to use reduction of order, but that is not as easy to follow.
So you get the first equations $$ 0=r(r-1)c_0\\ -c_0=(r+1)rc_1\\ -c_1=(r+2)(r+1)c_2 $$ etc. In the case $r=0$ the second equation tells you that $c_0=0$ and $c_1$ is arbitrary and determines the other coefficients, $$ y_0(x)=c_1(x-\frac12x^2\pm\dots). $$ In the case $r=1$ you get a solution based on $c_0$, but now with additional factor $x=x^r$ so that $$ y_1(x)=c_0x(1-\frac12x+\frac1{12}x^2\mp\dots) $$ which in the end is the same solution. You get the second basis solution by order reduction, set $y_2(x)=u(x)y_0(x)$ where $c_1=1$ with the equation $$ 0=x(u''y_0+2u'y_0')\implies u'(x)=y_0(x)^{-2}=x^{-2}\left(1 + x + \frac{7}{12}x^2 + \frac{19}{72}x^3 + \frac{149}{1440}x^4 + \dots\right) \\ \implies u(x)=-\frac1x+\ln(x)+\frac{7}{12}x + \frac{19}{144}x^2 + \frac{149}{4320}x^3 + \dots $$ which gives the solution that is singular in $x=0$.