11: If you tried the first approach and chose the square root with "plus", you obtained
This integral fits the box "ratio of linear functions", so use the appropriate substitution. Dividing by the root with "minus" would lead to a similar integral.
If you tried the second approach, you obtained
This is a typical "roots of quadratics" integral, so use the appropriate substitution. Since exactly this root was already integrated in the appropriate box, we will outline the calculations only briefly. It is also possible to use a hyperbolic substitution; this seems easier, the method is explained in the same "box".