Calculus 2
posted by Heather .
integral of sinh(p*ln(x/xsub0))dx

using a rather than x_{0} for ease of typing,
∫sinh(p ln(x/a)) dx
recall that sinh(u) = (e^u  e^u)/2
so, since
e^(p ln(x/a)) = (e^(ln(x/a)))^p = (x/a)^p
and, since e^u = 1/e^u,
e^(p ln(x/a)) = (a/x)^p
sinh(p ln(x/a)) = 1/2 (x^p/a^p  a^p/x^p)
integrate that to get
1/2 (x^(p+1)/((p+1)(a^p)) + a^p/(p1)x^(p1)))
(p1)x^(2p+1) + (p+1)xa^(2p)

2(p^21) a^p x^p
you can play around with that if you want. If you get really ambitious, you might even end up with
letting u = p*ln(x/a),
x(p cosh(u)  sinh(u))/(p^21)