# How do you find the integral of #dx/sqrt(x^2-4)#?

we can either use a trig substitution

this isa standard integral

substituting back

which can be simplified as required

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To find the integral of ( \frac{dx}{\sqrt{x^2 - 4}} ), you can use the trigonometric substitution method.

Let ( x = 2\sec(\theta) ), then ( dx = 2\sec(\theta)\tan(\theta) , d\theta ).

Substitute ( x = 2\sec(\theta) ) and ( dx = 2\sec(\theta)\tan(\theta) , d\theta ) into the integral:

[ \int \frac{dx}{\sqrt{x^2 - 4}} = \int \frac{2\sec(\theta)\tan(\theta)}{\sqrt{(2\sec(\theta))^2 - 4}} , d\theta ]

[ = \int \frac{2\sec(\theta)\tan(\theta)}{\sqrt{4\tan^2(\theta)}} , d\theta ]

[ = \int \frac{2\sec(\theta)\tan(\theta)}{2\tan(\theta)} , d\theta ]

[ = \int \sec(\theta) , d\theta ]

[ = \ln|\sec(\theta) + \tan(\theta)| + C ]

Substitute back ( \theta = \sec^{-1}\left(\frac{x}{2}\right) ):

[ = \ln\left|\sec\left(\sec^{-1}\left(\frac{x}{2}\right)\right) + \tan\left(\sec^{-1}\left(\frac{x}{2}\right)\right)\right| + C ]

[ = \ln\left|\frac{x}{2} + \sqrt{\frac{x^2}{4} - 1}\right| + C ]

Where ( C ) is the constant of integration.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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