`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`