Use the comparison test to determine whether the integral is convergent or divergent.

Question

$\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} $ )?

Answer 1

HINT

You should consider $$\int^\infty_1 \frac{x+1}{\sqrt{x^4-x}} dx=\int^2_1 \frac{x+1}{\sqrt{x^4-x}} dx+\int^\infty_2 \frac{x+1}{\sqrt{x^4-x}} dx$$

which diverges, indeed

$$\int^2_1 \frac{x+1}{\sqrt{x^4-x}} dx$$

converges but

$$\int^\infty_2 \frac{x+1}{\sqrt{x^4-x}} dx$$

diverges by comparison with $\frac1x$.

Answer 2

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.

Answer 3

A more "eye-balling" way to see it. For large $x$, you may approximate $$ x+1\approx x\\ \sqrt{x^4-x}\approx x^2 $$ meaning near infinity, your integral looks like $$ \int_{M}^\infty \frac{\mathrm dx}{x} $$ You can dress the above up in more rigor, and make sure it is right, but this is the intuition. Namely, $$ \frac{x+1}{\sqrt{x}\sqrt{x^3-1}}\stackrel{x+1\geq x}{\geq} \frac{\sqrt{x}}{\sqrt{x^3-1}}\\ \stackrel{\sqrt{x^3-1}\leq \sqrt{x^3}}{\geq}\frac{1}{x} $$ now use monotonicity of integration. Note also, as a bonus, your integral converges at $1$. Since the numerator is bounded near $1$, it won't hurt us and we can examine $$ \frac{1}{\sqrt{x^4-x}}=\frac{1}{\sqrt{x^3+x^2+x}}\frac{1}{\sqrt{x-1}} $$ again ignore the bounded pieces in the product and single out the singularity $$ \frac{1}{\sqrt{x-1}} $$ which is integrable near $1$ by the $p$ test.