Asked by Noel

intergral 2x^2/(x^2+4x+8)
Answered by Reiny
I did a long division and got

2x^2/(x^2+4x+8) = 2 - (4x+8)/(x^2 + 4x + 8)
= 2 - 2(2x+4)/(x^2 + 4x + 8)

I noticed that the derivative of x^2 + 4x + 8 is 2x+4 , which I have sitting on top.
Ahhh, logs!!!!

∫2x^2/(x^2+4x+8) dx
=∫ ( 2 - 2(2x+4)/x^2 + 4x + 8) ) dx
= 2x - 4 ln(x^2 + 4x + 8)
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