Math
posted by Ashwin .
Evaluate Integral x^2+4 / x^4+x^2+16 dx

gotta get creative with this one. Note that
x^4+x^2+16 = x^48x^2+16  9x^2 = (x^24)^2  (3x)^2
= (x^24)^2 [(3x/(x^24)^2)^2 + 1]
so, if you let u = 3x/(x^24)
du = 3(x^2+4)/(x^24)^2 and you have
integral 1/3 du/(1+u^2)
take it from there!
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