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Determine if $$\int_1 ^\infty \frac {dx}{x^2+x} $$ is divergent or convergent. If convergent: determine its value.

Tip: When $ x\ge1 $ is $ \frac 1 {x^2} \ge \frac 1 {x^2+x} = \frac 1 {x} - \frac {1} {x+1} $

Don't really know where to start here. Finding convergence/divergence really difficult so any tips on how to tackle questions like this one is appreciated.

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3 Answers 3

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You know the integral:

$$\int_1^\infty {1\over x^2}dx$$

Is convergent due to the p-series test. Then using the comparison test, we know that for all $x \geq 0$:

$${1\over x^2+x} \leq {1\over x^2}$$

Thus, the integral converges.

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HINT

Note that for $x\to \infty$

$$\frac {1}{x^2+x}\sim \frac{1}{x^2}$$

then use limit comparison test with $\int_1 ^\infty \frac {1}{x^2}dx$.

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  • $\begingroup$ Oh, so could one say that since we know that $\frac 1 {x^2} \ge \frac 1 {x^2+x} $ and $\frac 1 {x^2}$ converges we know that this improper integral converges? $\endgroup$
    – gbgult
    Commented Mar 5, 2018 at 13:37
  • $\begingroup$ yes of course by direct comparison or also that by limit comparison $\frac{\frac1{x^2+x}}{\frac1{x^2}}\to 1$ $\endgroup$
    – user
    Commented Mar 5, 2018 at 13:39
  • $\begingroup$ I've suggested limit comparison because it is more general and more easy to apply also for more difficult integrals, thus it is a good test to know; indeed not Always we can see an inequality as for the present case, while an asympthotic behaviour and a limit are more easy to apply in general $\endgroup$
    – user
    Commented Mar 5, 2018 at 13:40
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use that $$\int\frac{1}{x^2+x}dx=\ln\left(\frac{x}{x+1}\right)+C$$

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