Integrating with U-Substitution

Printable View

June 14th 2008, 09:12 AM
JonathanEyoon

Integrating 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)
June 14th 2008, 09:20 AM
Krizalid

$e^{-3\ln x}=e^{\ln x^{-3}}=\frac{1}{x^{3}},$ hence $\frac{x+1}{x^{3}}=\frac{1}{x^{2}}+\frac{1}{x^{3}}.$
June 14th 2008, 09:22 AM
JonathanEyoon

HUH?

How did it go from there to 1 / x^3? (Surprised)
June 14th 2008, 09:24 AM
Mathstud28

Quote:

Originally Posted by JonathanEyoon

HUH?

How did it go from there to 1 / x^3? (Surprised)

$\ln(e^{u(x)})=u(x)$ since they are inverse opeartions

so $-3\ln(x)=\ln(x^{-3})$

So we have that $e^{-3\ln(x)}=e^{\ln(x^{-3})}=x^{-3}=\frac{1}{x^3}$
June 14th 2008, 09:32 AM
JonathanEyoon

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.
June 14th 2008, 09:39 AM
Mathstud28

Quote:

Originally Posted by JonathanEyoon

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.

I do not know what you mean? How did you get $\ln(x^{-3})\ln(e)$

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

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

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

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

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

also note that u(x) is just any function
June 14th 2008, 09:40 AM
Silver

Quote:

Originally Posted by JonathanEyoon

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.

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 $e^{lnx^-3}$ , $e$ and $ln$ cancel and you end up with $x^{-3}$ which is the same as $\frac{1}{x^3}$
June 14th 2008, 09:44 AM
JonathanEyoon

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?
June 14th 2008, 09:46 AM
Mathstud28

Quote:

Originally Posted by JonathanEyoon

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?

(Clapping)

$e^{-\ln(x-1)}=e^{\ln\left(\frac{1}{x-1}\right)}=\frac{1}{x-1}$
June 14th 2008, 09:47 AM
Silver

Quote:

Originally Posted by JonathanEyoon

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?

Yes (Clapping)
June 14th 2008, 09:51 AM
JonathanEyoon

I appreciate the help guys~!