calculus
posted by Frank .
Integrate x^2+5x+1dx/(2x^3+15x^2+6x+23)^1/2

Let 2x^3+15x^2+6x+23 = u
du = 6x^2 + 30x + 6
= 6(x^2 + 5x + 1)dx
You got lucky with this one!
Therefore the integrand can be written
(1/6)du/u^1/2
and the integral is
(1/3)*u^1/2
= (1/3)(2x^3+15x^2+6x+23)^1/2
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