Calculus of a Single Variable, Chapter 5, 5.7, Section 5.7, Problem 8

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