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How can i find whether the integral

$$\int^{\infty}_{1}\frac{\sqrt{x}-1}{x^2+x+2}dx$$ in converge or diverges

What i try

I try to solve the Given Integal by substuting $x=t^2$ and $dx=2dt$ and changing limits

And it convert into $$2\int^{\infty}_{1}\frac{t^2-t}{t^4+t^2+2}dt$$

But solve it also very tedious job.

I have seems that we have to campare the original integral. But did not know how to campare it.

Could some help me. Thanks

jacky
  • 5,194

1 Answers1

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On the domain we have $$0 \le \frac{\sqrt{x}-1}{x^2+x+2} \le \frac{\sqrt{x}}{x^2} = \frac{1}{x^{1.5}}$$

Hence: $$0 \le \int^{\infty}_{1}\frac{\sqrt{x}-1}{x^2+x+2}dx \le \int^{\infty}_{1}\frac{1}{x^{1.5}}dx = 2$$ So the integral converges.

Gono
  • 5,598