Determining if integral converges or diverges

[tmt1]

$\int_{3}^{\infty} \frac{1}{\sqrt{x} - 1} \,dx$

I need to find if this converges or diverges.

I'm trying u-substitution, so $u = \sqrt{x} - 1$.

Therefore, $du = \frac{1}{2\sqrt{x}} dx$.

I'm not sure how to proceed from here.

 

[topsquark]

$\int_{3}^{\infty} \frac{1}{\sqrt{x} - 1} \,dx$

I need to find if this converges or diverges.

I'm trying u-substitution, so $u = \sqrt{x} - 1$.

Therefore, $du = \frac{1}{2\sqrt{x}} dx$.

I'm not sure how to proceed from here.

Good so far. Just keep going:
\(\displaystyle \sqrt{x} = u + 1\)

Thus
\(\displaystyle dx = 2(u + 1) ~ du\)

etc.

-Dan

 

[MarkFL]

I would continue as follows:

\(\displaystyle dx=2\sqrt{x}\,du=(2u+2)\,du\)

And so the integral becomes:

\(\displaystyle I=\int_{\sqrt{3}-1}^{\infty}\frac{2u+2}{u}\,du=2\int_{\sqrt{3}-1}^{\infty} 1+\frac{1}{u}\,du=2\lim_{t\to\infty}\left(\int_{\sqrt{3}-1}^{t} 1+\frac{1}{u}\,du\right)\)

 

[Greg]

$$\int_3^\infty\frac{1}{\sqrt{x}-1}\text{ d}x$$

$$t^2=x,\quad2t\text{ d}t=\text{d}x$$

$$2\int_\sqrt{3}^\infty\frac{t}{t-1}\text{ d}t=2\int_\sqrt{3}^\infty1+\frac{1}{t-1}\text{ d}t$$

etc.