dx
(x − p)n √(ax2 + 2bx + c)

can be deduced by differentiating (n − 1) times with respect to p.

(iii.)

(Hx + K) dx
(αx2 + 2βx + γ) √(ax2 + 2bx + c)

can be reduced by the substitution y2 = (ax2 + 2bx + c)/(αx2 + 2βx + γ) to the form

A ∫ dy+ B ∫ dy
√(λ1 − y2) √(y2 − λ2)

where A and B are constants, and λ1 and λ2 are the two values of λ for which (a − λα)x2 + 2(b − λβ)x + c − λγ is a perfect square (see A. G. Greenhill, A Chapter in the Integral Calculus, London, 1888).

(iv.) ƒxm (axn + b)p dx, in which m, n, p are rational, can be reduced, by putting axn = bt, to depend upon ƒtq (1 + t)p dt. If p is an integer and q a fraction r/s, we put t = us. If q is an integer and p = r/s we put 1 + t = us. If p + q is an integer and p = r/s we put 1 + t = tus. These integrals, called “binomial integrals,” were investigated by Newton (De quadratura curvarum).

(v.)

dx = log tan x,
sin x 2