int t / sqrt(1-t^4) dt Find the indefinite integral

Indefinite integral are written in the form of int f(x) dx = F(x) +C
 where: f(x) as the integrand
           F(x) as the anti-derivative function 
           C  as the arbitrary constant known as constant of integration
 
In the given problem: int t/sqrt(1-t^4)dt , we follow: int f(t)dt =F(t) +C.
The problem can be rewritten as:
int (t *dt)/sqrt(1^2-(t^2)^2)
 
This resembles the basic integration formula for inverse sine function:
int (du)/sqrt(a^2-u^2) = arcsin(u/a) +C
Using u-substitution, we let u = t^2 then du = 2t*dt or (du)/2= t*dt .
Note: a^2 = 1 then a = 1
The indefinite integral will be:
int (t *dt)/sqrt(1^2-(t^2)^2)=int ((du)/2)/sqrt(1^2-(u)^2)
Applying the basic property of integration: int c*f(x)dx = c int f(x) dx , we get:
(1/2) int (du)/sqrt(1^2-u^2)
Applying the basic integral formula for inverse sine function:
(1/2) int (du)/sqrt(1^2-u^2)=1/2arcsin(u/1) +C
                                    =1/2arcsin(u)+C
Plug-in  u=t^2  in 1/2arcsin(u) +C to express the indefinite intergral in terms of  int f(t)dt=F(t)+C :
int t/sqrt(1-t^4)dt =1/2arcsin(t^2) +C