# How do you integrate # (x^3)((x^2 + 4)^(1/2)) dx#?

Use the substitution x^2+4=t^2#,

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To integrate ( \int x^3 \sqrt{x^2 + 4} , dx ), use the substitution method.

Let: [ u = x^2 + 4 ] [ \frac{du}{dx} = 2x ] [ du = 2x , dx ]

From the given integral, we can express ( x^3 , dx ) as ( \frac{1}{2} du ).

Substituting these values into the integral:

[ \int x^3 \sqrt{x^2 + 4} , dx = \frac{1}{2} \int u^{1/2} , du ]

Integrate ( u^{1/2} ) with respect to ( u ):

[ \int u^{1/2} , du = \frac{2}{3} u^{3/2} + C ]

Substitute back for ( u ):

[ \frac{2}{3} (x^2 + 4)^{3/2} + C ]

Thus, the integral of ( \int x^3 \sqrt{x^2 + 4} , dx ) is ( \frac{2}{3} (x^2 + 4)^{3/2} + 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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