# How do you find the integral of # x^3/(1+x^2)#?

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To find the integral of ( \frac{x^3}{1+x^2} ), you can use a substitution method. Let ( u = 1 + x^2 ). Then ( du = 2x , dx ). Rearranging for ( dx ), we get ( \frac{du}{2x} = dx ).

Substituting ( u ) and ( dx ) into the integral:

[ \int \frac{x^3}{1+x^2} , dx = \int \frac{x^3}{u} \cdot \frac{du}{2x} ]

[ = \frac{1}{2} \int \frac{x^2}{u} , du ]

[ = \frac{1}{2} \int \frac{x^2}{1+x^2} , du ]

Now, use another substitution: ( v = 1 + x^2 ), so ( dv = 2x , dx ). This time we need to express ( x^2 ) in terms of ( v ). Since ( v = 1 + x^2 ), ( x^2 = v - 1 ).

Substituting ( v ) and ( x^2 ) into the integral:

[ \frac{1}{2} \int \frac{x^2}{1+x^2} , du = \frac{1}{2} \int \frac{v - 1}{v} , dv ]

[ = \frac{1}{2} \int \left( 1 - \frac{1}{v} \right) , dv ]

[ = \frac{1}{2} \left( v - \ln|v| \right) + C ]

Substitute back ( v = 1 + x^2 ) to get the final result:

[ = \frac{1}{2} \left( 1 + x^2 - \ln|1+x^2| \right) + C ]

So, the integral of ( \frac{x^3}{1+x^2} ) is ( \frac{1}{2} \left( 1 + x^2 - \ln|1+x^2| \right) + 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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