# How do you find the integral #intsqrt(x^2-1)/xdx# ?

The answer is

We need

Perform this integral by substitution

Therefore,

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To find the integral ( \int \frac{\sqrt{x^2 - 1}}{x} , dx ), you can use trigonometric substitution. Let ( x = \sec(\theta) ), then ( dx = \sec(\theta) \tan(\theta) , d\theta ). Substituting these into the integral gives:

[ \int \frac{\sqrt{x^2 - 1}}{x} , dx = \int \frac{\sqrt{\sec^2(\theta) - 1}}{\sec(\theta)} \sec(\theta) \tan(\theta) , d\theta ]

Simplify the expression under the square root:

[ \sqrt{\sec^2(\theta) - 1} = \sqrt{\tan^2(\theta)} = \tan(\theta) ]

So the integral becomes:

[ \int \tan^2(\theta) , d\theta ]

Now, you can integrate ( \tan^2(\theta) ) using trigonometric identities. After integration, substitute back ( x = \sec(\theta) ) to get the final result.

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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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