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confirm that F(x)= (1/4)x^4lin(x) - (-1/16)x^4+12 is the anti-derivative for F(x)=x^3ln(x) for values of x>0

 Nov 27, 2014

Best Answer 

 #2
avatar+33603 
+10

Integrate directly using integration by parts:

integrate by parts

k = 12 is just one possibility, so it's more accurate to say that x4ln(x)/4 - x4/16 + 12 is an antiderivative rather than the antiderivative.

Notice that Melody is correct to suggest the sign of the second term in the original question is wrong.

.

 Nov 27, 2014
 #1
avatar+118587 
+5

confirm that F(x)= (1/4)x^4lin(x) - (-1/16)x^4+12 is the anti-derivative for F(x)=x^3ln(x) for values of x>0

 

$$\\F(x)=\frac{x^4ln(x)}{4}-\;\frac{-1x^4}{16}+12\\\\
F(x)=\frac{x^4ln(x)}{4}\;+\;\frac{x^4}{16}+12\\\\
F'(x)=\frac{4x^3ln(x)+\frac{x^4}{x}}{4}\;+\;\frac{4x^3}{16}\\\\
F'(x)=\frac{4x^3ln(x)+x^3}{4}\;+\;\frac{1x^3}{4}\\\\
F'(x)=\frac{4x^3ln(x)+2x^3}{4}\\\\
F'(x)=\frac{2x^3ln(x)+x^3}{2}\\\\
F'(x)=x^3ln(x)+\frac{x^3}{2}$$

 

If the original question had (+1/16) instead of (-1/16) it would have worked.

Then again, maybe I did something wrong  

 Nov 27, 2014
 #2
avatar+33603 
+10
Best Answer

Integrate directly using integration by parts:

integrate by parts

k = 12 is just one possibility, so it's more accurate to say that x4ln(x)/4 - x4/16 + 12 is an antiderivative rather than the antiderivative.

Notice that Melody is correct to suggest the sign of the second term in the original question is wrong.

.

Alan Nov 27, 2014

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