Determine if improper integral is convergent or divergent

Question

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.

Answer 1

HINT

Note that for $x\to \infty$

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

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

Answer 2

use that $$\int\frac{1}{x^2+x}dx=\ln\left(\frac{x}{x+1}\right)+C$$

Answer 3

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.