I've spent about a good 30 minutes on this problem and I can't seem to find a way to use u-sub to make this problem work

(x + 1)e^(-3lnx)

Any help is appreciated! (Crying)

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- Jun 14th 2008, 09:12 AMJonathanEyoonIntegrating with U-Substitution
I've spent about a good 30 minutes on this problem and I can't seem to find a way to use u-sub to make this problem work

(x + 1)e^(-3lnx)

Any help is appreciated! (Crying) - Jun 14th 2008, 09:20 AMKrizalid
$\displaystyle e^{-3\ln x}=e^{\ln x^{-3}}=\frac{1}{x^{3}},$ hence $\displaystyle \frac{x+1}{x^{3}}=\frac{1}{x^{2}}+\frac{1}{x^{3}}.$

- Jun 14th 2008, 09:22 AMJonathanEyoon
HUH?

How did it go from there to 1 / x^3? (Surprised) - Jun 14th 2008, 09:24 AMMathstud28
- Jun 14th 2008, 09:32 AMJonathanEyoon
Ok i think I see what happened here but I have a question.

When it got to

e^ln(x^-3)

What you did was bring the ln(x^-3) down by taking the ln to do so right? But wouldn't that end up being

ln(x^-3) lne since you have to do ln of something to bring down whatever is in the exponent? Sorry for the stupid questions =/ just really want to understand it. - Jun 14th 2008, 09:39 AMMathstud28
I do not know what you mean? How did you get $\displaystyle \ln(x^{-3})\ln(e)$

Let $\displaystyle f(x)$ be some functions, and let $\displaystyle g(x)=f^{-1}(x)$ or the inverse, then by definition

$\displaystyle f(g(x))=x\Rightarrow{f(g(u(x)))=u(x)}$

So seeing that $\displaystyle f(x)=e^x$ and $\displaystyle g(x)=\ln(x)$

we would have $\displaystyle f(g(u(x)))=u(x)$

or more particularly $\displaystyle f(g(x^{-3}))=x^{-3}$

also note that u(x) is just any function - Jun 14th 2008, 09:40 AMSilver
think of 'natural log cancelling with e' when nautral log is the power of e, because they are the inverse of eachother.

:)

so when it is $\displaystyle e^{lnx^-3}$, $\displaystyle e$ and $\displaystyle ln$ cancel and you end up with $\displaystyle x^{-3}$ which is the same as $\displaystyle \frac{1}{x^3}$ - Jun 14th 2008, 09:44 AMJonathanEyoon
OoHhhH I totally forgot that e and ln were inverse functions!! *sigh* I must remember this!!

So for instance another question that is similar to the one posted above is

(x^2 + 2x + 1)e^(-ln(x - 1)) dx

is equal to

(x^2 + 2x + 1) / (x - 1) dx

right? - Jun 14th 2008, 09:46 AMMathstud28
- Jun 14th 2008, 09:47 AMSilver
- Jun 14th 2008, 09:51 AMJonathanEyoon
I appreciate the help guys~!