So, doing the substitution x = 4 sin(t), we also know that dx = 4 cos(t) dt

So: (x^2) / (16 - x^2) dx --> 16 (sin(t))^2 / (16 - 16(sin(t))^2) * 4 cos(t) dt

--> 16 (sin(t))^2 / 16(1 - (sin(t))^2) * 4 cos(t) dt --> (sin(t))^2 / (cos(t))^2 * 4 cos(t) dt

--> 4 dt * (1 - (cos(t))^2) / cos(t) --> 4 sec(t) dt - 4 cos(t) dt

Now integrate:

4 * ln | sec(t) + tan(t) | - 4 * sin(t) + C

Now go back and substitute t = arcsin(x/4):

4 * ln | (4 + x) / sqrt(16 - x^2) | - x + C

Which can be further reduced to: ln | (4 + x)^4 / (16 - x^2)^2 | - x + C

--> ln | (4 + x)^4 / [(4 - x)^2 * (4 + x)^2] | - x + C --> ln | (4 + x)^2 / (4 - x)^2 | - x + C

--> 2 ln | (4 + x) / (4 - x) | - x + C

Use substitution and we have:

(16sin^2(t))dt/(16 - 16sin^2(t))

Now we take out 16

16(sin^2(t))dt/(1 - sin^2(t))

sin^2(x) + cos^2(x) = 1 thus 1 - sin^2(x) = cos^2(x)

16(sin^2(t))dt/cos^2(t)

Simplify

16tan^2(t)dt

You can integrate from there, I don't remember the integral for tan^2(x) off the top of my head.