# Small indefinite integral mistake

• February 10th 2010, 02:16 PM
s3a
Small indefinite integral mistake
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!
• February 10th 2010, 02:30 PM
Keithfert488
$\frac{\mathrm{d} \ln (8x)}{\mathrm{d} x}\neq \frac{8}{x}$

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