tan^(1/3)x dx

Let tanx=u^3

=>(sec x)^2 dx=3u^2 du

now, (sec x)^2=1+ (tan x)^2=1+u^6

=>dx=[(3u^2)/(1+ u^6)] du

therefore, ?tan^(1/3)x dx=3?[u X u^2/(1+u^6)]du

=>3?[u^3/(1+u^6)]du
let u^2=t
=>(2u) du=dt

=>3?[u^3/(1+u^6)]du=(3/2)?[t/(...

Now, partial fractions will be used
we know that, t/(1+t^3)=t/(1+t)(1+t^2-t)....... a^3 + b^3=(a+b)(a^2 + b^2 + ab)

let, t/(1+t)(1+t^2-t)=A/(1+t) + (Bt + C)/(1+t^2-t)

=>t/(1+t)(1+t^2-t)=[A(1+t^2 -t) + (Bt + C)(1+t)]/(1+t)(1+t^2 -t)................taking LCM

equating coefficients of u, u^2, constants on both sides
=>1=-A+B+C...................e... coefficients of u
0= A+B .......................equatin... coefficients of u^2
0= A+C .......................equatin... coefficients of constant
=>A=-1/3, B=1/3, C=1/3

=> (3/2)?[t/(1+t^3)]dt=
(3/2)?[(1/3)/(1+t)]dt + (3/2)?[(1/3)(1+t)/(1+t^2-t)]dt


=>(-1/2)log|1+t| + (1/2?[(1+t)/(1+t^2-t)]dt

=>(-1/2)log|1+t| + (1/4)?[(2+2t)/(1+t^2-t)]dt

we know, d/dt(1+ t^2 - t)=2t -1
=>(-1/2)log|1+t| + (1/4?[(2+2t+1-1)/(1+t^2-t)]dt

=>(-1/2)log|1+t| + (1/4)log|1+t^2 -t| + (3/4)?1/(1+t^2-t)dt

we know, 1+t^2-t=(t-1/2)^2 +3/4

=>(-1/2)log|1+t| + (1/4)log|1+t^2 -t| + (3/4)?1/[(t-1/2)^2 + 3/4]dt

=>(-1/2)log|1+t| + (1/4)log|1+t^2 -t| +
(3/4)1/(2/sqrt(3) arctan[2(t-1/2)/sqrt(3)


substituting original values t=u^2

=>(-1/2)log|1+u^2| + (1/4)log|1+u^4 -u^2| +
(3/4)1/(2/sqrt(3) arctan[2(u^2-1/2)/sqrt(3)

substituting original values u=(tanx)^1/3

=>(-1/2)log|1+(tanx)^2/3| + (1/4)log|1+(tanx)^4/3 -(tanx)^2/3| +
(3/4)1/(2/sqrt(3) arctan[2((tanx)^2/3 - 1/2)/sqrt(3) + C

where arctan(x) means tan inverse x
results used=
1)?1/(1+x)dx= log|1+x| + c
2)?1/(a^2+x^2)dx=
(1/a)arctan(x/a) + C...........'a' is a constant