^{-1}(x) has for its derivative 1/√[x

^{2}-1], so if we get something that looks similar in form (by a u-substitution, for example), we can integrate.

^{2}-5] is almost there, but we need to do something about the 5 (based on the derivative, we need a 1 there). If we factor out the 5 from the quadratic, we can recover the 1:

^{2}/5-1)]

^{2}/5-1] dx)

^{2}/5 together into a single term squared:

^{2}/5 = (x/√5)

^{2}

^{2}-1 ] ==> (√5 du)/√[u

^{2}-1]

^{-1}(u)=cosh

^{-1}(x/√5).

^{-1}at the two limits of integration, you can substitute in the ln(...) definition for evaluation instead (since they differ by a constant, the constant will disappear when you subtract). Just remember to carry over the x/√5 when you go from cosh

^{-1}to ln