Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 2

In this case, this would be x = 2tan(t).
This is because the the trigonometric identity tan^2(t) + 1 = sec^2(t) can then be applied:
sqrt(4 + x^2) = sqrt(4 + 4tan^2(t)) = sqrt(4(tan^2(t) + 1))
= sqrt(4sec^2(t)) = 2sec(t)
Also, if x = 2tan(t), then dx = 2sec^2(t)dt and x^3 = 8tan^3(t)dt
Plugging all this into original integral, we get
int (8tan^3(t))/(2sec(t)) 2sec^2(t)dt
This simplifies to
int (8tan^3(t))/cos(t) dt
Rewriting tangent as tan(t) = sin(t)/cos(t) , we get
int (8sin^3(t))/(cos^4(t)) dt
Now we can rewrite sin^3(t) as sin^2(t) * sin(t) = (1 - cos^2(t)) sin(t)
and use substitution:
u = cos(t)
du = -sin(t)dt
Then integral becomes
int (8(1 - u^2))/u^4 (-du) = int (8(u^2 - 1))/u^4 du
which can be broken up into two integrals of power functions:
int u^2/u^4 du = int u^(-2) du = - 1/u + C_1
and int 1/u^4 du= int u^(-4) du = -1/(3u^3) + C_2
Then the original integral will be, if we subtract the results and combine the constants into one:
-8/u + 8/(3u^3) + C
Now recall that u = cos(t) and x = 2tan(t)
Use the Pythageorean identity again:
tan^2(t) + 1 = sec^2(t)
x^2/4 + 1 = 1/u^2
From here, u = 2/sqrt(x^2 + 4) . Plugging this into our result for the integral, we get
-4sqrt(x^2 + 4) + 1/3(x^2 + 4)sqrt(x^2 + 4) + C .
This is the final answer.