The question is asking to prove:

$\displaystyle \int^0_x\frac{1}{\sqrt{t^2-a^2})}=ln(x+\sqrt{x^2+a^2})-ln(a)$

So I think I understand when it is said:

$\displaystyle L'(x)=\frac{1}{\sqrt{a^2+x^2}}$

so that must be because the definite integral is from 0 to x so L'=f(x) or another way of saying this would be that t was replaced by x and the resulting function is L'.

In regards to R'(x) I see $\displaystyle ln(x+\sqrt{x^2+a^2})-ln(a)$ as $\displaystyle F(b)-F(a)$ per the first fundamental theorem of calculus. So then it is explained:

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Since L'(x)=R'(x), we have L(x)=R(x)+C for some constant C=L(x)-R(x). The constant C may be evaluated by assigning a value to x; the most convienient choice is x=0 which gives:

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The part above I find I am right now struggling to understand what is meant by this. L'(x)=R'(x). I thought the definite integral was = to F(b)-F(a). I went through the attached lecture notes before I tried the exercise and I think I understand a good amount of those notes although the proof of FTC1 by FTC2 is a little bit confusing to me the rest I think I was able to grasp what was being said.

Thank You!