Question

Integral of 1/(u^3 + 3u) du, without using partial fractions? Many thanks!

Answer 1

This can be done by some none obvious algebra manipulation:
integral [ 1/(u^3 + 3u) du ]
= 1/3 * integral [ 3/(u^3 + 3u) du ]
=1/3 * integral [ 3/{u*(u^2+3) du ]
=1/3 * integral [ {3+u^2-u^2} / {u*(u^2+3)} du ]
=1/3*integral [ {3+u^2} / {u*(u^2+3)} du ] - 1/3*integral [ {u^2} / {u*(u^2+3)} du ]
=1/3*integral [ 1/u du ] - 1/3*integral [ u/{u^2+3} du]
let u^2+3 = t => dt = 1/(2*u) du
=1/3*ln(u) - 1/3 * 1/2 * integral (1/t dt)
=1/3*ln(u) - 1/3*1/2*ln(t)
=1/3*ln(u) - 1/3*ln(sqrt(t))
=1/3*ln( u / sqrt(t) )
=1/3*ln( u / sqrt(u^2+3) ) + C