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Superposition and Method of Images

I'm stuck on part (b) of the following question. I've found the formula for the electric field by multiplying the equation from part (a) by the resistivity, which I thought I could integrate to find the voltage which I could then use to find the resistance.

I've looked up a solution for a similar question (although in that one you only need to find the voltage) which, when it integrates the bracketed fractions, ends up with:

1(y−b2)+1(y+b2)\frac{1}{(y-\frac{b}{2})}+\frac{1}{(y+\frac{b}{2})}

Essentially just getting rid of the squares. I don't understand how they did that. The limits they were using were from infinity to b/2+a, which I'm assuming are the same as I need to use.

(edited 9 years ago)

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