cosh-1(x) has for its derivative 1/√[x2-1], so if we get something that looks similar in form (by a u-substitution, for example), we can integrate.
1/√[x2-5] 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:
And in fact, we can take √5 out of the whole denominator now. Because it's just a constant, we can take it out of the entire integral:
Next we want to pull the x2/5 together into a single term squared:
x2/5 = (x/√5)2
This is important because now we can make a simple u-sub and get rid of it entirely:
The integral is transformed into
dx/√[ (x/√5)2-1 ] ==> (√5 du)/√[u2-1]
The √5 we pulled out earlier (in the denominator) and the √5 we created with the u-substitution (in the numerator) cancel each other out. The integral we're left with will exactly give us cosh-1(u)=cosh-1(x/√5).
When you evaluate cosh-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