Calculating Integral: \int\frac{x^3dx}{\sqrt{2-x}} Solution and Alternatives

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SUMMARY

The integral \(\int\frac{x^3dx}{\sqrt{2-x}}\) can be solved using the substitution \(v=2-x\), leading to the expression \(-\int\frac{(2-v)^3dv}{\sqrt{v}}\). This method simplifies the integral into manageable parts: \(-\int\frac{8dv}{\sqrt{v}} + \int 12\sqrt{v}dv - \int 6v\sqrt{v}dv - \int v^2\sqrt{v}dv\). An alternative approach involves substituting \(v^2\) for \(2-x\), resulting in a radical-free expression that can be integrated directly.

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Homework Statement



[tex]\int\frac{x^3dx}{\sqrt{2-x}}[/tex]




The Attempt at a Solution


I solved it in this way
for v=2-x
[tex]\int-\frac{(2-v)^3dv}{\sqrt{v}}=-\int\frac{(8-12v+6v^2+v^3)dv}{\sqrt{v}}<br /> =-\int\frac{8dv}{\sqrt{v}}+\int 12\sqrt{v}dv-\int 6v\sqrt{v}dv-\int v^2\sqrt{v}dv[/tex]
is it true?
Is there any other methods?
 
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That's one way of doing it. If you substitute v2 for 2-x, you will have a radical-free expression to integrate.
 
Ok,very nice.
 

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