How do you use comparison test to determine is the integral is convergent or divergent given $\int \frac{x}{8 x^{2} + 2 x^{2} - 1} d x$ $int x / (8x^2 + 2x^2 - 1) dx$ ?

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Answer 1 · 2015-05-14T12:08:58

First of all a think there is a mistake: maybe $8 x^{2}$ $8x^2$ would be $8 x^{3}$ $8x^3$ ,

so the function is:

$\frac{x}{8 x^{3} + 2 x^{2} - 1} ~ \frac{1}{8 x^{2}}$ $x/(8x^3+2x^2-1)~1/(8x^2)$ ( $~ =$ $~ =$ asymptotic)

that is a convergent function because it's a generalized armonic series, $\frac{1}{x^{α}}$ $1/x^alpha$ , with $α > 1$ $alpha>1$ .

How do you use comparison test to determine is the integral is convergent or divergent given ∫x8x2+2x2−1dxint x / (8x^2 + 2x^2 - 1) dx ?