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
Integrate directly using integration 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.
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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
$$\\F(x)=\frac{x^4ln(x)}{4}-\;\frac{-1x^4}{16}+12\\\\
F(x)=\frac{x^4ln(x)}{4}\;+\;\frac{x^4}{16}+12\\\\
F'(x)=\frac{4x^3ln(x)+\frac{x^4}{x}}{4}\;+\;\frac{4x^3}{16}\\\\
F'(x)=\frac{4x^3ln(x)+x^3}{4}\;+\;\frac{1x^3}{4}\\\\
F'(x)=\frac{4x^3ln(x)+2x^3}{4}\\\\
F'(x)=\frac{2x^3ln(x)+x^3}{2}\\\\
F'(x)=x^3ln(x)+\frac{x^3}{2}$$
If the original question had (+1/16) instead of (-1/16) it would have worked.
Then again, maybe I did something wrong
Integrate directly using integration 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.
.