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Given this integral: $$ \int_1^\infty {x+1\over\sqrt{x^4-x}}dx $$

Determine whether this integral is convergent or divergent.

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I know it will be divergent, therefore I must compare it to a function that is smaller than the given integral for all x values AND is also divergent. What is an integral that satisfies this? –  EngGenie Jan 20 '13 at 21:51

2 Answers 2

up vote 4 down vote accepted

Hint :

$$\int_1^\infty \frac{x+1}{\sqrt{x^4-x}}dx \geq \int_1^\infty \frac{x+1}{\sqrt{x^4}}dx=\int_1^\infty \frac{x+1}{x^2}dx$$ $$=\int_1^\infty \frac{x}{x^2}dx +\int_1^\infty \frac{1}{x^2}dx=\int_1^\infty \frac{1}{x}dx +\int_1^\infty \frac{1}{x^2}dx$$

Make sure to justify why each step is correct!

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It's straightforward if you resort to the inequality

$$\frac{1}{x}\le{x+1\over\sqrt{x^4-x}}$$

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