Small indefinite integral mistake

**s3a** · February 10th 2010, 01:16 PM

Here is what wolfram alpha says:
http://www.wolframalpha.com/input/?i...al+2%2F(xln(8x))

and the attached pdf file is what I did.

What I do not get is why you don't divide it by 8 like I multiplied by x for the du/dx algebraic manipulation.

Any help would be greatly appreciated!
Thanks in advance!

**Keithfert488** · February 10th 2010, 01:30 PM

$\frac{\mathrm{d} \ln (8x)}{\mathrm{d} x}\neq \frac{8}{x}$

$\frac{\mathrm{d} \ln (8x)}{\mathrm{d} x}=\frac{{u}'}{u}$
where u=8x
Thus,

$\frac{\mathrm{d} \ln (8x)}{\mathrm{d} x}=\frac{8}{8x}=\frac{1}{x}$

You might want to try it again with that.

The reason that $\frac{\mathrm{d} \ln (8x)}{\mathrm{d} x}$ is the same as $\frac{\mathrm{d} \ln (x)}{\mathrm{d} x}$
is because $\ln(8x)=\ln(8)+\ln(x)$ due to the properties of logs. So ln(8x) is a constant added to ln(x)

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