Integration involving e and derivative of ln

**thebacontalks2me** · May 6th 2009, 09:45 PM

Find the indefinite integral.

I know you have to use U substitution U=1/x^5 but I get lost from there.

Find the derivative of y with respect to x.

I was told to use the product rule on this one but again I'm stuck.

Use implicit differentiation to find dy / dx.

Lost on how to find the derivative of ln(xy)

thanks in advance for any help.

**chisigma** · May 6th 2009, 10:41 PM

The only you need in last question is to remember that $\ln (xy) = \ln x + \ln y$ ...

Kind regards

$\chi$ $\sigma$

**warmonk102** · May 6th 2009, 10:43 PM

for the first one u=1/x^5 then du=-5/x^6 dx so can you see it now?

**chisigma** · May 6th 2009, 10:52 PM

... for the first one is...

$u= x^{-5} \rightarrow x=u^{-\frac{1}{5}} \rightarrow dx= -\frac{1}{5}\cdot u^{-\frac{6}{5}}\cdot du$

... after that...

Kind regards

$\chi$ $\sigma$

**thebacontalks2me** · May 6th 2009, 11:03 PM

I got du= -5/x^6dx. I assume you divide by -5 so you end up with -1/5du=1/x^6. so you end up with -1/5 e^u du. Dunno what to do from there.

**chisigma** · May 6th 2009, 11:27 PM

... the substitution $u=x^{-5} \rightarrow dx= -\frac{1}{5}\cdot u^{-\frac{6}{5}}\cdot du$ leads to the substitution...

$\int \frac{e^{x^{-5}}}{x^{6}}\cdot dx \rightarrow -\frac{1}{5}\cdot \int e^{u}\cdot du$

Kind regards

$\chi$ $\sigma$

**thebacontalks2me** · May 6th 2009, 11:42 PM

thanks for the help.

just need help on that one.

**chisigma** · May 7th 2009, 12:03 AM

If $f=u\cdot v$ then $df= u\cdot dv + v\cdot du$ . So that...

Kind regards

$\chi$ $\sigma$

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