how do you show that intergrate from 0 to infinity of

1/ (x^2 + sq root x) dx converges?

i tried to do comparison test by saying that it is less than integrate from 0 to infinity of

1/ sq root x but it doesnt show that it converges.

Printable View

- May 5th 2011, 03:20 AMalexandrabel90converges
how do you show that intergrate from 0 to infinity of

1/ (x^2 + sq root x) dx converges?

i tried to do comparison test by saying that it is less than integrate from 0 to infinity of

1/ sq root x but it doesnt show that it converges. - May 5th 2011, 03:33 AMFernandoRevilla
If

*f*(*x*) is the integrand function, we have

$\displaystyle \displaystyle\lim_{x \to{+}\infty}{\frac{f(x)}{1/x^2}}=\ldots=1\neq 0$

$\displaystyle \displaystyle\lim_{x \to 0^+}{\frac{f(x)}{1/\sqrt{x}}}=\ldots=1\neq 0$

So,

$\displaystyle \int_0^1 f(x)dx\;,\quad \int_1^{+\infty} f(x)dx$

are convergent