# How do you integrate #(1/(25 + x^2) ) dx#?

We can rewrite this as

By definition, we know that the integration of the inverse of a sum of squares results in a trigonometric function as follows:

Applying this formula to your function, we have this:

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To integrate ( \frac{1}{25 + x^2} ), we can use the arctangent substitution method. Let ( x = 5\tan(\theta) ). Then ( dx = 5\sec^2(\theta) d\theta ).

Substituting into the integral, we get: [ \int \frac{1}{25 + x^2} dx = \int \frac{1}{25 + 25\tan^2(\theta)} \cdot 5\sec^2(\theta) d\theta ] [ = \int \frac{1}{25(1 + \tan^2(\theta))} \cdot 5\sec^2(\theta) d\theta ] [ = \int \frac{1}{25\sec^2(\theta)} \cdot 5\sec^2(\theta) d\theta ] [ = \int 1 d\theta ] [ = \theta + C ]

Now, we need to find ( \theta ). We know that ( x = 5\tan(\theta) ), so ( \tan(\theta) = \frac{x}{5} ).

[ \theta = \arctan\left(\frac{x}{5}\right) ]

So, the final answer is: [ \int \frac{1}{25 + x^2} dx = \arctan\left(\frac{x}{5}\right) + C ]

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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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