∫√(x3-1)dx.
The square root only exists for x≥1. x3-1=x3(1-1/x3).
We can express √(x3-1) as x3/2(1-1/x3)½ and this can be expressed by a series:
x3/2{1-½/x3-(½)(½)(1/x6)(½)-(½)(½)(3/2)(1/x9)(1/3!)-(½)(½)(3/2)(5/2)(1/x12)(1/4!)-...}=
x3/2-(½)x-3/2-(⅛)x-9/2-(1/16)x-15/2-(5/128)x-21/2-(7/256)x-27/2-...,
which can be written: x3/2-(½)x-3/2-∑(1×...×(2n-3)/(2nn!)x(3-6n)/2 for n≥2.
When this is integrated we get:
⅖x5/2+x-½+x-7/2/28+x-13/2/104+...+C.
This can be expressed as the infinite series:
C+⅖x5/2+x-½+∑(2/(6n-5))(1×...×(2n-3)/(2nn!)x(5-6n)/2 for n≥2.