tony873004

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[tex]

\int_1^9 {\frac{1}{{2x}}\,dx}

[/tex]

[tex]

F(x) = \frac{{\ln \left| x \right|}}{2}

[/tex]

[tex]

\frac{{\ln \left| 9 \right|}}{2} - \frac{{\ln \left| 1 \right|}}{2} = \frac{{\ln \left| 9 \right|}}{2} = \ln 3

[/tex]

The answer ln3 came from the back of the book. I realize from using my calculator that ln9 / 2 = ln3 , but I'm not sure why. I guess I forgot my rules of ln.

Also, since the anti-derivate section gives 1/x as ln abs(x), was I correct in carrying the absolute value brackets to the ln abs(9) / 2 ? Why did the back of the book drop the absolute value brackets from the answer?

Thanks!!

\int_1^9 {\frac{1}{{2x}}\,dx}

[/tex]

[tex]

F(x) = \frac{{\ln \left| x \right|}}{2}

[/tex]

[tex]

\frac{{\ln \left| 9 \right|}}{2} - \frac{{\ln \left| 1 \right|}}{2} = \frac{{\ln \left| 9 \right|}}{2} = \ln 3

[/tex]

The answer ln3 came from the back of the book. I realize from using my calculator that ln9 / 2 = ln3 , but I'm not sure why. I guess I forgot my rules of ln.

Also, since the anti-derivate section gives 1/x as ln abs(x), was I correct in carrying the absolute value brackets to the ln abs(9) / 2 ? Why did the back of the book drop the absolute value brackets from the answer?

Thanks!!

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