интеграл dx/x*(1+x^2)^1/3

Cначала домножим числ и знам на х
Int= 1/2 Int (2x dx)/{x^2 * (1+x^2)^1/3}= 1/2 Int d(x^2)/{x^2 * (1+x^2)^1/3}
теперь делаем замену x^2=t
1/ 2 Int dt/ {t * (1+t)^1/3} теперь делаем замену 1+t=z^3 dt=3z^2 dz
= 3/2 Int z dz/(z^3-1)= 3/2 Int z dz /(z-1)(z^2+z+1) =
= 3/2 Int {1/3 * [1/(z-1) + (1-z)/(z^2+z+1)] } dz =
= 1/2 {Int dz /(z-1) + Int (1-z)/(z^2+z+1) dz }=
= 1/2 ln |z-1| + Int dz/(z^2+z+1) - Int z dz / (z^2+z+1)=
z^2+z+1= (z+1/2)^2 + 3/4 u=z+1/2 a=(sqrt3)/2
Int dz/(z^2+z+1)= Int d(z+1/2) / [(z+1/2)^2 +a^2] =
= 1/a * arctg U/a + const= 2 (sqrt3)/3 * arctg [2u (sqrt3)/3} + const=
= 2 (sqrt3)/3 * arctg [2*(z+1/2)* (sqrt3)/3} + const=
= 2 (sqrt3)/3 * arctg [2*( (1+t)^1/3 +1/2)* (sqrt3)/3} + const=
2 (sqrt3)/3 * arctg [2*( (1+x^2)^1/3 +1/2)* (sqrt3)/3} + const=

Int z dz / (z^2+z+1) = Int (z+1/2) dz / [(z+1/2)^2+3/4]=
= 1/2 Int d(u^2) / [ u^2 + 3/4 ]= 1/2 Int d(u^2+ 3/4) / [ u^2 + 3/4 ]=
= 1/2 ln (u^2 + 3/4) + const =1/2 ln (z^2+z+1) + const=
= 1/2 ln [ (1+t)^2/3 + (1+t)^1/3 +1 ] + const=
= 1/2 ln [ (1+x^2)^2/3 + (1+x^2)^1/3 +1 ] + const
Осталось собрать три части в один ответ.

по частям
а затем замена переменной

http://www.exponenta.ru/EDUCAT/class/courses/ma/theme15/theory.asp