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This almost seems to easy to be true. Is there anything wrong with this:

[itex]\mbox{Find} \int_0^1 x^p\, (ln\, {x})^3 \, dx [/itex]

[itex]\mbox{First let } F(p) = \int_0^1 x^p\,dx = \frac{x^{p+1}}{p+1} |_0^1 = \frac{1}{p+1}[/itex]

[itex]\frac{\partial}{\partial p}( \frac{1}{p+1}) = \frac{-1}{(p+1)^2}[/itex]

[itex]\mbox{then }F'(p) = \int_0^1 x^p\,ln\,x\,dx = \frac{-1}{(p+1)^2}[/itex]

[itex]\mbox{so now let } G(p) = \int_0^1 x^p\,ln\,x\,dx = \frac{-1}{(p+1)^2}[/itex]

[itex]\frac{\partial}{\partial p} (\frac{-1}{(p+1)^2}) = \frac{2}{(p+1)^3}[/itex]

[itex]\mbox{then }G'(p) = \int_0^1 x^p\,(ln\,x)^2\,dx = \frac{2}{(p+1)^3} [/itex]

[itex]\mbox{Finally let } H(p) = \int_0^1 x^p\,(ln\,x)^2\,dx = \frac{2}{(p+1)^3} [/itex]

[itex]\frac{\partial}{\partial p}( \frac{2}{(p+1)^3}) = \frac{-6}{(p+1)^4}[/itex]

[itex]\mbox{therefore } H'(p) = \int_0^1 x^p\,(ln\,x)^3\,dx = \frac{-6}{(p+1)^4}[/itex]

[itex]\mbox{Find} \int_0^1 x^p\, (ln\, {x})^3 \, dx [/itex]

[itex]\mbox{First let } F(p) = \int_0^1 x^p\,dx = \frac{x^{p+1}}{p+1} |_0^1 = \frac{1}{p+1}[/itex]

[itex]\frac{\partial}{\partial p}( \frac{1}{p+1}) = \frac{-1}{(p+1)^2}[/itex]

[itex]\mbox{then }F'(p) = \int_0^1 x^p\,ln\,x\,dx = \frac{-1}{(p+1)^2}[/itex]

[itex]\mbox{so now let } G(p) = \int_0^1 x^p\,ln\,x\,dx = \frac{-1}{(p+1)^2}[/itex]

[itex]\frac{\partial}{\partial p} (\frac{-1}{(p+1)^2}) = \frac{2}{(p+1)^3}[/itex]

[itex]\mbox{then }G'(p) = \int_0^1 x^p\,(ln\,x)^2\,dx = \frac{2}{(p+1)^3} [/itex]

[itex]\mbox{Finally let } H(p) = \int_0^1 x^p\,(ln\,x)^2\,dx = \frac{2}{(p+1)^3} [/itex]

[itex]\frac{\partial}{\partial p}( \frac{2}{(p+1)^3}) = \frac{-6}{(p+1)^4}[/itex]

[itex]\mbox{therefore } H'(p) = \int_0^1 x^p\,(ln\,x)^3\,dx = \frac{-6}{(p+1)^4}[/itex]

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