Solving an Improper Integral Homework Equation

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Bashyboy
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Homework Statement


[itex]\int \frac{dx}{\sqrt{x^2-4}}[/itex]


Homework Equations





The Attempt at a Solution



I tried trig-substitution, by realizing that [itex]cot\theta = \frac{4}{\sqrt{x^2-4}}[/itex]

and that [itex]-4sin\theta = dx[/itex]

My answer, though, found after the substitution and integration, is very different from the books: mine is [itex]- \frac{\sqrt{x^2-4}}{x}[/itex], theirs is [itex]ln|x+\sqrt{x^2-4}|[/itex]

How do you account for this variation?
 
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Bashyboy said:

Homework Statement


[itex]\int \frac{dx}{\sqrt{x^2-4}}[/itex]


Homework Equations





The Attempt at a Solution



I tried trig-substitution, by realizing that [itex]cot\theta = \frac{4}{\sqrt{x^2-4}}[/itex]

and that [itex]-4sin\theta = dx[/itex]
You have mistakes in your substitution. One of the legs in your triangle should be 2, not 4. Also, your equation for dx is incorrect.
Bashyboy said:
My answer, though, found after the substitution and integration, is very different from the books: mine is [itex]- \frac{\sqrt{x^2-4}}{x}[/itex], theirs is [itex]ln|x+\sqrt{x^2-4}|[/itex]

How do you account for this variation?