Divergence of $\int_8^\infty\frac{\sqrt{x+1}}{x+\sqrt{x}} dx$

Question

Determine wheter the following integral will converge.

$$\int_8^\infty\frac{\sqrt{x+1}}{x+\sqrt{x}} dx$$

Can i go like this?

$\frac{\sqrt{x+1}}{x+\sqrt{x}} \geq \frac{{1}}{x+\sqrt{x}} \geq \frac{1}{x+x} \geq \frac{1}{2x}$

and now $\frac{1}{2}\int_8^M\frac{1}{x}dx$, which will diverge since $p \leq 1$?

Answer 1

Yes, you are right.

Also, $$\frac{\sqrt{x+1}}{x+\sqrt{x}}>\frac{1}{2\sqrt{x}}$$ and since $$\int\limits_{8}^{+\infty}\frac{1}{2\sqrt{x}}=\lim_{x\rightarrow+\infty}(\sqrt{x}-\sqrt8)=+\infty,$$ we obtain the same result.

Answer 2

What you did is correct. You can also use the fact that$$\lim_{x\to\infty}\frac{\frac{\sqrt{x+1}}{x+\sqrt x}}{\frac1{\sqrt x}}=\lim_{x\to\infty}\frac{\sqrt{x^2+x}}{x+\sqrt x}=\lim_{x\to\infty}\sqrt{\frac{x^2+x}{x^2+2x\sqrt x+x}}=1$$and therefore, since the integral $\int_8^\infty\frac1{\sqrt x}\,\mathrm dx$ diverges, your integral diverges too.