test the integral $\int_{0}^{\infty} \frac {x}{3x^4 + 5x^2 +1}dx$ for convergence.
My thought
Can I compare it with 1/(3x^4)?
Any hints for the solution are appreciated!
2026-03-30 03:03:27.1774839807
test the integral $\int_{0}^{\infty} \frac {x}{3x^4 + 5x^2 +1}dx$ for convergence
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$I=\int_0^\infty \frac x{3x^4 + 5x^2 +1}dx$
$\forall x>0, 0\leq\frac x{3x^4 + 5x^2 +1}\leq\frac x{x^4}=\frac1{x^3}$
$\implies 0\leq I=\int_0^1 \frac x{3x^4 + 5x^2 +1}dx+\int_1^\infty \frac x{3x^4 + 5x^2 +1}dx\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \leq\int_0^1 \frac x{3x^4 + 5x^2 +1}dx+\int_1^\infty \frac 1{x^3}dx\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int_0^1 \frac x{3x^4 + 5x^2 +1}dx+\int_1^\infty \frac 1{x^3}dx\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int_0^1 \frac x{3x^4 + 5x^2 +1}dx+\frac12$
Since $\frac x{3x^4 + 5x^2 +1}$ is continuous on $[0,1]$ (the denominator has no real zeroes), it is bounded there and so is the proper integral $\int_0^1 \frac x{3x^4 + 5x^2 +1}dx$.