Integrating with U-Substitution

**JonathanEyoon** · June 14th 2008, 09:12 AM

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!

**Krizalid** · June 14th 2008, 09:20 AM

$e^{-3\ln x}=e^{\ln x^{-3}}=\frac{1}{x^{3}},$ hence $\frac{x+1}{x^{3}}=\frac{1}{x^{2}}+\frac{1}{x^{3}}.$

**JonathanEyoon** · June 14th 2008, 09:22 AM

HUH?

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

**Mathstud28** · June 14th 2008, 09:24 AM

Originally Posted by JonathanEyoon

HUH?

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

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

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

So we have that $e^{-3\ln(x)}=e^{\ln(x^{-3})}=x^{-3}=\frac{1}{x^3}$

**JonathanEyoon** · June 14th 2008, 09:32 AM

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.

**Mathstud28** · June 14th 2008, 09:39 AM

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

**Silver** · June 14th 2008, 09:40 AM

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 $\frac{1}{x^3}$

**JonathanEyoon** · June 14th 2008, 09:44 AM

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?

**Mathstud28** · June 14th 2008, 09:46 AM

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?

$e^{-\ln(x-1)}=e^{\ln\left(\frac{1}{x-1}\right)}=\frac{1}{x-1}$

**Silver** · June 14th 2008, 09:47 AM

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

**JonathanEyoon** · June 14th 2008, 09:51 AM

I appreciate the help guys~!

Thread: Integrating with U-Substitution

LinkBack

Thread Tools

Display

Integrating with U-Substitution

Similar Math Help Forum Discussions

Integrating Cosine and sine integrals by substitution

Integrating by Polar Substitution?

Integrating correct substitution,,

Integrating 'sec(x)' Using T-Substitution

Integrating using Trig Substitution

Search Tags