# How do you find the antiderivative of #int 1/(x^2+10x+29) dx#?

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To find the antiderivative of (\int \frac{1}{x^2+10x+29} , dx), you can complete the square in the denominator to rewrite the expression:

[ x^2 + 10x + 29 = (x + 5)^2 + 4 ]

Now, substitute (u = x + 5), then (du = dx):

[ \int \frac{1}{(x + 5)^2 + 4} , dx = \int \frac{1}{u^2 + 4} , du ]

Now, you can use a trigonometric substitution. Let (u = 2 \tan(\theta)), then (du = 2 \sec^2(\theta) , d\theta). Rewrite the integral in terms of (\theta):

[ \int \frac{1}{u^2 + 4} , du = \frac{1}{2} \int \frac{1}{\tan^2(\theta) + 1} \cdot 2 \sec^2(\theta) , d\theta ]

This simplifies to:

[ \frac{1}{2} \int \sec^2(\theta) , d\theta = \frac{1}{2} \tan(\theta) + C ]

Now, you need to back-substitute for (x):

[ \frac{1}{2} \tan(\theta) + C = \frac{1}{2} \tan\left(\arctan\left(\frac{x+5}{2}\right)\right) + C ]

[ = \frac{1}{2} \left(\frac{x+5}{2}\right) + C = \frac{1}{4} (x + 5) + C = \frac{x}{4} + \frac{5}{4} + C ]

So, the antiderivative of (\frac{1}{x^2+10x+29}) is (\frac{x}{4} + \frac{5}{4} + 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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