$\int^\infty_1 \frac{x+1}{\sqrt{x^4-x}} dx $
So far I have found that $\frac{x+1}{\sqrt{x^4-x}} \ge \frac{1}{x^4} $ and then since $ \frac{1}{x^4} $ converges (known by the P-test) I know that the integral converges as well. But the answer in the textbook is that the integral diverges.
I am not sure what I am missing.
I was also wondering if there are any tricks for picking the function to compare the initial integral (in this case $ \frac{1}{x^4} $ )?
Best Answer
Note that $$ \frac{x+1}{\sqrt{x^4-x}}\geq \frac{x+1}{\sqrt{x^4}}\ge\frac{x}{x^2}=1/x. $$ Now apply the comparison test on $[2,\infty]$. Alternatively observe that $$ \frac{x+1}{\sqrt{x^4-x}}\sim\frac{1}{x} $$ and apply the limit comparison test.