Integrals Inequallity: A <= int_1^{infinity} (2 + sin(x))/(1 + x^2)^{1.5} dx <= B

[AlexSendler100%]

[math]A \leq \int_1^{\infty} \frac{2+\sin x}{\left(1+x^2\right)^{1.5}} \quad d x \leq B \quad[/math]
A and B are positive now I tried the comprehension test. I have got that two of the sides converge but how do I find the value itself??

But in the end, I GOT a and b that are negative using p series and that could not be true because they should be positive so what I do wrong???

 

[BigBeachBanana]

Omitting the limits for brevity.

[math]\int\dfrac{1}{(2x^2)^{1.5}}\, dx \leq \int\dfrac{1}{(1+x^2)^{1.5}}\, dx \leq \int \frac{2+\sin x}{\left(1+x^2\right)^{1.5}} \,dx \leq \int \frac{3}{\left(1+x^2\right)^{1.5}} \, dx \leq \int \frac{3}{(x^2)^{1.5}}\, dx[/math]
Or

[math]\int\dfrac{1}{(2x^2)^{1.5}}\, dx \leq \int \frac{2+\sin x}{\left(1+x^2\right)^{1.5}} \,dx \leq \int \frac{3}{(x^2)^{1.5}}\, dx[/math]
[math]\dfrac{1}{2\sqrt{2}} \int_1^\infty \dfrac{1}{x^{1.5}} \, dx \leq \int_1^\infty \frac{2+\sin x}{\left(1+x^2\right)^{1.5}} \,dx \leq \int_{1}^\infty \frac{3}{x^3}\, dx[/math]
[math]\dfrac{1}{\sqrt{2}} \leq \int_1^\infty \frac{2+\sin x}{\left(1+x^2\right)^{1.5}} \,dx \leq\dfrac{3}{2} [/math]