intergral $\int _{ 1 }^{ \infty }{ \sqrt { \frac { x+2 }{ { x }^{ 4 }-1 } } dx } $

Question

This question has been killing me,

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

Show that this is either divergent or convergent

Answer 1

Near $1^+$,

$$x^4-1=(x^2+1)(x+1)(x-1) $$

hence

$$\sqrt {\frac {x+2}{x^4-1}}\sim \sqrt { \frac {3}{4 (x-1)}} $$

thus it converges near $ 1^+$ because $\int_1\frac {dx}{\sqrt {x-1}} $ converges.

As it converges near $+\infty $, (see DeepSea's above answer), it is convergent.

Answer 2

hint: $\sqrt{\dfrac{x+2}{x^4+1}}\approx x^{-\frac{3}{2}}$. Use the $p$ series test with $p > 1$.

Answer 3

$$\int _{ 1 }^{ \infty }\frac{\sqrt(x+2)}{\sqrt(x^4 -1)}dx < \int _{ 1 }^{ \infty }\frac{\sqrt(4x)}{\sqrt(\frac{1}{4}x^4 )}dx = \int _{ 1 }^{ \infty }\frac{4}{x^{\frac{3}{2}}}dx$$

Which converges. Therefore by comparison the integral convergences.