# How do you evaluate the integral #int x^2/(16+x^2)#?

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To evaluate the integral (\int \frac{x^2}{16+x^2} , dx), we use the method of partial fractions.

First, we write the integrand as (\frac{x^2}{16+x^2} = \frac{A}{4+x} + \frac{B}{4-x}) where A and B are constants to be determined.

Next, we find common denominators and equate coefficients to get (x^2 = A(4-x) + B(4+x)).

Solving for A and B, we find (A = 1) and (B = 1).

Now, we rewrite the integral as (\int \frac{1}{4+x} + \frac{1}{4-x} , dx).

Integrating term by term, we get (\int \frac{1}{4+x} , dx + \int \frac{1}{4-x} , dx = \ln|4+x| - \ln|4-x| + C), where C is the constant of integration.

So, (\int \frac{x^2}{16+x^2} , dx = \ln|\frac{4+x}{4-x}| + 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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