How do you show whether the improper integral $\int \frac{d x}{x^{2} + sin x}$ $int dx / (x^2 + sinx)$ converges or diverges from 2 to infinity?

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Answer 1 · 2016-07-19T17:41:31

$\int_{2}^{\infty} \frac{d x}{x^{2} + sin x} \leq {log}_{e} (\sqrt{3})$ $int_2^oo dx / (x^2 + sinx) lelog_e(sqrt(3))$

$\int \frac{d x}{x^{2} + sin x} \leq \int \frac{d x}{x^{2} - 1}$ $int dx / (x^2 + sinx)le int dx/(x^2-1)$

so

$int_2^oo dx / (x^2 + sinx) lelim_{x->oo}log_e sqrt(abs(1-x)/abs(1+x)) +log_e(sqrt(3))$

so

$int_2^oo dx / (x^2 + sinx) lelog_e(sqrt(3))$

How do you show whether the improper integral int dx / (x^2 + sinx)∫dxx2+sinxint dx / (x^2 + sinx) converges or diverges from 2 to infinity?