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

Answer 1

Ezekiel Alexander

#=(1/15)(x^2+4)^(3/2)(3x^2+7)+C#

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

Then 2x dx =2t dt. So, #dx=(t/x)dt=t/sqrt((t^2-4)) dt#

Now, #int x^3sqrt(x^2+4) dx =int (t^2-4)^(3/2)t^2/sqrt(t^2-4)dt#

#=int(t^2-4)t^2dt=int(t^4-4t^2)dt#

#=t^5/5-t^3/3+ C#

#=t^3/15(3t^2-5)+C#

#=(1/15)(x^2+4)^(3/2)(3(x^2+4)-5)+C#

#=(1/15)(x^2+4)^(3/2)(3x^2+7)+C#

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Answer 2

Aurora Alvarado

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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Answer from HIX Tutor

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.