`int 1 / (xsqrt(x^4-4)) dx` Find the indefinite integral

For the given integral: `int 1/(xsqrt(x^4-4))dx` , we may apply u-substitution by letting:

`u =x^4-4 ` then ` du = 4x^3 dx` .

Rearrange `du = 4x^3 dx` into `(du)/( 4x^3)= dx`

Plug-in `u =x^4-4`  and `(du)/( 4x^3)= dx` , we get:

`int 1/(xsqrt(x^4-4))dx =int 1/(xsqrt(u))* (du)/( 4x^3)`

                       ` =int 1/(4x^4sqrt(u))du`

Recall `u =x^4-4` then adding 4 on both sides becomes: `u + 4 = x^4` .

Plug-in `x^4...

For the given integral: `int 1/(xsqrt(x^4-4))dx` , we may apply u-substitution by letting:

`u =x^4-4 ` then ` du = 4x^3 dx` .

Rearrange `du = 4x^3 dx` into `(du)/( 4x^3)= dx`

Plug-in `u =x^4-4`  and `(du)/( 4x^3)= dx` , we get:

`int 1/(xsqrt(x^4-4))dx =int 1/(xsqrt(u))* (du)/( 4x^3)`

                       ` =int 1/(4x^4sqrt(u))du`

Recall `u =x^4-4` then adding 4 on both sides becomes: `u + 4 = x^4` .

Plug-in `x^4 =u+4` in the integral:

`int 1/(4x^4sqrt(u))du` =`int 1/(4(u+4)sqrt(u))du`

Apply the basic integration property: `int c*f(x) dx = c int f(x) dx` :

`int 1/(4(u+4)sqrt(u))du=1/4int 1/((u+4)sqrt(u))du`

Apply another set of substitution by letting:

`v =sqrt(u) `  which is the same as `v^2 =u` .

Then taking the derivative on both sides, we get `2v dv = du` .

Plug-in `u =v^2` , `du = 2v dv` , and `sqrt(u)=v `  , we get:

`1/4 int 1/((u+4)sqrt(u))du = 1/4int 1/((v^2+4)v)(2v dv)`

We simplify by cancelling out common factors v and 2:

`1/4int 1/((v^2+4)v)(2v dv) =1/2int (dv)/(v^2+4) or1/2int (dv)/(v^2+2^2)`  

The integral part resembles the integration formula:

`int (du)/(u^2+a^2) = (1/a) arctan (u/a) +C`

 Then, 

`1/2 int (dv)/(v^2+4) =1/2 *(1/2) arctan (v/2) +C`

                        ` =1/4 arctan (v/2) +C`

Recall that we let `v =sqrt(u) ` and `u =x^4-4 `  then  ` v = sqrt(x^4-4)`

Plug-in `v = sqrt(x^4-4)` in  `1/4 arctan (v/2) +C`  to get the final answer:

`int 1/(xsqrt(x^4-4))dx =1/4 arctan (sqrt(x^4-4)/2) +C`