OMG integrals...!?
Answer:
If u = sqr(x), du = -1/(2*sqr(x))=-1/(2*u)
So the integral becomes,
INT(-2*u/(2+u) du)
You can find this in a table of integrals. It evaluates to:
-2*ln(-4+x)+2*sqr(x)+2*ln((-4-...
Well, if we do u^2=x, then our integral change to 1/(2+u). And this one can be treated as f'(u)/f(u), so the the primitive of that equation is ln(2+u). Now, you just have to rewrite u and apply the limits of integration (if were present).
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