cosh^{1}(x) has for its derivative 1/√[x^{2}1], so if we get something that looks similar in form (by a usubstitution, for example), we can integrate.
1/√[x^{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:
1/√[5(x^{2}/51)]
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:
(1/√5)*Int(1/√[x^{2}/51] dx)
Next we want to pull the x^{2}/5 together into a single term squared:
x^{2}/5 = (x/√5)^{2}
This is important because now we can make a simple usub and get rid of it entirely:
u=x/√5
√5 du=dx
The integral is transformed into
dx/√[ (x/√5)^{2}1 ] ==> (√5 du)/√[u^{2}1]
The √5 we pulled out earlier (in the denominator) and the √5 we created with the usubstitution (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
May 3

Dom V.