As mentioned in comment, this ODE is Legendre differential equation for l=1

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\left({1}-{x}^{{2}}\right)}{\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}\right]}+{l}{\left({l}+{1}\right)}{y}{\left({x}\right)}={0}\)

The standard Frobenius method will give us the Legendre polynomials.

By inspection, it is sort of obvious \(\displaystyle{y}{\left({x}\right)}={x}\) is one solution for this ODE. In general, if we already know one solution to a second order homogenous ODE, we don't need Frobenius method to find the other one.

Given any second order homogenous ODE in its Sturm-Liouville form

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{p}{\left({x}\right)}{\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}\right]}+{q}{\left({x}\right)}{y}{\left({x}\right)}={0}\)

Let \(\displaystyle{y}_{{1}}{\left({x}\right)}\) be a solution we already knew, let \(\displaystyle{y}_{{2}}{\left({x}\right)}\) be another solution we seek. We have

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{p}{\left({y}_{{1}}{\frac{{{\left.{d}{y}\right.}_{{2}}}}{{{\left.{d}{x}\right.}}}}-{y}_{{2}}{\frac{{{\left.{d}{y}\right.}_{{1}}}}{{{\left.{d}{x}\right.}}}}\right)}\right]}\)

\(\displaystyle={y}_{{1}}{\left[{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{p}{\frac{{{\left.{d}{y}\right.}_{{2}}}}{{{\left.{d}{x}\right.}}}}\right]}+{q}{y}_{{2}}\right]}-{y}_{{2}}{\left[{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{p}{\frac{{{\left.{d}{y}\right.}_{{1}}}}{{{\left.{d}{x}\right.}}}}\right]}+{q}{y}_{{1}}\right]}={0}\)

This implies

\(\displaystyle{p}{\left({x}\right)}{\left({y}_{{1}}{\left({x}\right)}{\frac{{{\left.{d}{y}\right.}_{{2}}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}-{y}_{{2}}{\left({x}\right)}{\frac{{{\left.{d}{y}\right.}_{{1}}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}\right)}={c}{o}{n}{s}{t}.\)

Let say's we fix the constant at the left to 1, we can rewrite above as

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\frac{{{y}_{{2}}{\left({x}\right)}}}{{{y}_{{1}}{\left({x}\right)}}}}\right]}={\frac{{{1}}}{{{p}{\left({x}\right)}{{y}_{{1}}^{{2}}}{\left({x}\right)}}}}\Rightarrow{y}_{{2}}{\left({x}\right)}\)

\(\displaystyle={y}_{{1}}{\left({x}\right)}{\left[\int^{{x}}{\frac{{{\left.{d}{t}\right.}}}{{{p}{\left({t}\right)}{y}_{{1}}{\left({t}\right)}^{{2}}}}}\right]}+{c}{o}{n}{s}{t}{]}\)

is another solution. Apply this to our ODE, the other solution we seek can be chosen as

\(\displaystyle{x}\int^{{x}}{\frac{{{1}}}{{{\left({1}-{t}^{{2}}\right)}{t}^{{2}}}}}{\left.{d}{t}\right.}={x}\int^{{x}}{\left[{\frac{{{1}}}{{{t}^{{2}}}}}+{\frac{{{1}}}{{{2}}}}{\left({\frac{{{1}}}{{{1}-{t}}}}+{\frac{{{1}}}{{{1}+{t}}}}\right)}\right]}{\left.{d}{t}\right.}\)

\(\displaystyle={x}{\left[-{\frac{{{1}}}{{{x}}}}+{\frac{{12}}{{\log{{\left({\frac{{{1}+{x}}}{{{1}-{x}}}}\right)}}}}}\right]}\)

\(\displaystyle={x}{{\text{tanh}}^{{-{1}}}{\left({x}\right)}}-{1}\)

The one metioned in Claude Leibovici's answer.

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\left({1}-{x}^{{2}}\right)}{\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}\right]}+{l}{\left({l}+{1}\right)}{y}{\left({x}\right)}={0}\)

The standard Frobenius method will give us the Legendre polynomials.

By inspection, it is sort of obvious \(\displaystyle{y}{\left({x}\right)}={x}\) is one solution for this ODE. In general, if we already know one solution to a second order homogenous ODE, we don't need Frobenius method to find the other one.

Given any second order homogenous ODE in its Sturm-Liouville form

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{p}{\left({x}\right)}{\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}\right]}+{q}{\left({x}\right)}{y}{\left({x}\right)}={0}\)

Let \(\displaystyle{y}_{{1}}{\left({x}\right)}\) be a solution we already knew, let \(\displaystyle{y}_{{2}}{\left({x}\right)}\) be another solution we seek. We have

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{p}{\left({y}_{{1}}{\frac{{{\left.{d}{y}\right.}_{{2}}}}{{{\left.{d}{x}\right.}}}}-{y}_{{2}}{\frac{{{\left.{d}{y}\right.}_{{1}}}}{{{\left.{d}{x}\right.}}}}\right)}\right]}\)

\(\displaystyle={y}_{{1}}{\left[{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{p}{\frac{{{\left.{d}{y}\right.}_{{2}}}}{{{\left.{d}{x}\right.}}}}\right]}+{q}{y}_{{2}}\right]}-{y}_{{2}}{\left[{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{p}{\frac{{{\left.{d}{y}\right.}_{{1}}}}{{{\left.{d}{x}\right.}}}}\right]}+{q}{y}_{{1}}\right]}={0}\)

This implies

\(\displaystyle{p}{\left({x}\right)}{\left({y}_{{1}}{\left({x}\right)}{\frac{{{\left.{d}{y}\right.}_{{2}}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}-{y}_{{2}}{\left({x}\right)}{\frac{{{\left.{d}{y}\right.}_{{1}}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}\right)}={c}{o}{n}{s}{t}.\)

Let say's we fix the constant at the left to 1, we can rewrite above as

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\frac{{{y}_{{2}}{\left({x}\right)}}}{{{y}_{{1}}{\left({x}\right)}}}}\right]}={\frac{{{1}}}{{{p}{\left({x}\right)}{{y}_{{1}}^{{2}}}{\left({x}\right)}}}}\Rightarrow{y}_{{2}}{\left({x}\right)}\)

\(\displaystyle={y}_{{1}}{\left({x}\right)}{\left[\int^{{x}}{\frac{{{\left.{d}{t}\right.}}}{{{p}{\left({t}\right)}{y}_{{1}}{\left({t}\right)}^{{2}}}}}\right]}+{c}{o}{n}{s}{t}{]}\)

is another solution. Apply this to our ODE, the other solution we seek can be chosen as

\(\displaystyle{x}\int^{{x}}{\frac{{{1}}}{{{\left({1}-{t}^{{2}}\right)}{t}^{{2}}}}}{\left.{d}{t}\right.}={x}\int^{{x}}{\left[{\frac{{{1}}}{{{t}^{{2}}}}}+{\frac{{{1}}}{{{2}}}}{\left({\frac{{{1}}}{{{1}-{t}}}}+{\frac{{{1}}}{{{1}+{t}}}}\right)}\right]}{\left.{d}{t}\right.}\)

\(\displaystyle={x}{\left[-{\frac{{{1}}}{{{x}}}}+{\frac{{12}}{{\log{{\left({\frac{{{1}+{x}}}{{{1}-{x}}}}\right)}}}}}\right]}\)

\(\displaystyle={x}{{\text{tanh}}^{{-{1}}}{\left({x}\right)}}-{1}\)

The one metioned in Claude Leibovici's answer.