A Very Nice Integral Problem From Russia

[@giuseppemalaguti435]

t=x^(1/6)...=>I=2t^3-3t^2+6t-6ln(1+t)=2√x-3x^(1/3)+6x^(1/6)-6ln(1+x^(1/6))+c...dovrei ricontrollare...

[@gelbkehlchen]

Solution:
∫1/(√x+∛x)*dx =
------------------
Substitution:
u = x^(1/6) x = u^6 √x = x^(3/6) = u³ ∛x = x^(2/6) = u²
du = 1/6*x^(-5/6)*dx dx = 6*x^(5/6)*du = 6*u^5*du
------------------
= 6*∫u^5/(u³+u²)*du = 6*∫u³/(u+1)*du =
------------------
Substitution: t = u+1 u = t-1 dt = du
------------------
= 6*∫(t-1)³/t*dt = 6*∫(t-1)*(t²-2t+1)/t*dt = 6*∫(t³-3t²+3t-1)/t*dt
= 6*∫(t²-3t+3-1/t)*dt = 6*(t³/3-3t²/2+3t-ln|t|+C)
= 2t³-9t²+18t-6*ln|t|+6C = 2*(u+1)³-9*(u+1)²+18*(u+1)-6*ln|u+1|+D
= 2*(u³+3u²+3u+1)-9*(u²+2u+1)+18*(u+1)-6*ln|u+1|+D
= 2u³+6u²+6u+2-9u²-18u-9+18u+18-6*ln|u+1|+D
= 2u³-3u²+6u-6*ln|u+1|+D+11
= 2*√x-3*∛x+6*x^(1/6)-6*ln|x^(1/6)+1|+E