# How do you find the antiderivative of #int x^3/(4+x^2) dx#?

Splitting up the integral:

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To find the antiderivative of ( \int \frac{x^3}{4+x^2} , dx ), you can use substitution. Let ( u = 4 + x^2 ). Then ( du = 2x , dx ). Rearranging, ( \frac{1}{2} du = x , dx ). Now substitute these into the integral:

( \int \frac{x^3}{4+x^2} , dx = \int \frac{x^2 \cdot x}{4+x^2} , dx )

Substitute ( u = 4 + x^2 ) and ( \frac{1}{2} du = x , dx ):

( = \frac{1}{2} \int \frac{x^2}{u} , du )

Now, we can split ( \frac{x^2}{u} ) into ( \frac{x^2}{4} \cdot \frac{1}{u} ):

( = \frac{1}{2} \cdot \frac{1}{4} \int \frac{x^2}{\frac{1}{4}(4+x^2)} , du )

Now, we have a simpler integral:

( = \frac{1}{8} \int \frac{1}{u} , du )

Integrate ( \frac{1}{u} ):

( = \frac{1}{8} \ln|u| + C )

Finally, substitute back ( u = 4 + x^2 ):

( = \frac{1}{8} \ln|4 + x^2| + C )

So, the antiderivative of ( \int \frac{x^3}{4+x^2} , dx ) is ( \frac{1}{8} \ln|4 + x^2| + 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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