MHB 242.q2.3a Int_2^4 dx/[x( ln x)^2]

[karush]

$\Large{242.Q2.3a} \\
\text{find the integral} \\
\displaystyle
\int_2^4 \frac{dx}{x( ln\, x)^2} $

 

[Greg]

Re: 242.q2.3a Int_2^4 dx/(x ln x)^2 dx

Where did you get this question?

 

[topsquark]

Re: 242.q2.3a Int_2^4 dx/(x ln x)^2 dx

$\Large{242.Q2.3a} \\
\text{find the integral} \\
\displaystyle
\int_2^4 \frac{dx}{x( ln\, x)^2} dx$
Yes this has 2 dx so ??

One problem is that you have too many "dx"s. :)

Hint: Do you know of a relationship between ln(x) and 1/x?

-Dan

 

[soroban]

Re: 242.q2.3a Int_2^4 dx/(x ln x)^2 dx

$\displaystyle \text{Find the integral: } \; \int_2^4 \frac{dx}{x(\ ln x)^2} $

$\displaystyle \text{We have: }\;\int (\ln x)^{-2\,}\frac{dx}{x} $

$\text{Let }u \,=\,\ln x \quad\Rightarrow\quad du \,=\,\frac{dx}{x}$

$\displaystyle \text{Substitute: }\;\int u^{-2}du$

$\text{Got it?}$

 

[karush]

Re: 242.q2.3a Int_2^4 dx/(x ln x)^2 dx

$\Large{242.Q2.3a} \\
\text{find the integral} $
$$\displaystyle
I_{3a}=\int_2^4 \frac{dx}{x( ln\, x)^2}=\ln\left({2}\right) \\
u=\ln\left({x}\right) \ \ \ du=\frac{1}{x}\, dx $$
$\text{then} $
$$\displaystyle
I_{3a}
=\int_2^4\frac{1}{{u}^{2}}\, du
= \left[-\frac{1}{u}\right]_2^4$$
So far??

 

[MarkFL]

Re: 242.q2.3a Int_2^4 dx/(x ln x)^2 dx

When you use a substitution in a definite integral, you need to not only rewrite the integrand and differential in terms of the new variable, but you also need to change the limits of integration in accordance with your substitution. :)

 

[karush]

Re: 242.q2.3a Int_2^4 dx/(x ln x)^2 dx

ok i keep forgetting to do so

 

[topsquark]

Re: 242.q2.3a Int_2^4 dx/(x ln x)^2 dx

$\Large{242.Q2.3a} \\
\text{find the integral} $
$$\displaystyle
I_{3a}=\int_2^4 \frac{dx}{x( ln\, x)^2}=\ln\left({2}\right) \\
u=\ln\left({x}\right) \ \ \ du=\frac{1}{x}\, dx $$
$\text{integrate and back substitute } u=ln(x)$
$$\displaystyle
I_{3a}
=\int_{\ln{2}}^{2\ln{2}}\frac{1}{{u}^{2}}\, du
= \left[-\frac{1}{\ln{x}}\right]
_{\ln\left({2}\right)}^{2\ln\left({2}\right)}$$

You didn't back substitute in you last step. It's [math]\left[-\frac{1}{\ln{x}}\right]
_{2}^{4}[/math]

-Dan

 

[karush]

Re: 242.q2.3a Int_2^4 dx/(x ln x)^2 dx

$\Large{242.Q2.3a} \\
\text{find the integral} $
$$\displaystyle
I_{3a}=\int_2^4 \frac{dx}{x( ln\, x)^2}=\ln\left({2}\right) \\
u=\ln\left({x}\right) \ \ \ du=\frac{1}{x}\, dx $$
$\text{back substitute } u=ln(x)$
$$\displaystyle
I_{3a}
=\int_{\ln{2}}^{\ln{4}}\frac{1}{{u}^{2}}\, du
= \left[-\frac{1}{\ln{x}}\right]
_{2}^{4}=
\frac{1}{\ln 2}-\frac{1}{\ln 4}$$

 

[MarkFL]

Re: 242.q2.3a Int_2^4 dx/(x ln x)^2 dx

$\Large{242.Q2.3a} \\
\text{find the integral} $
$$\displaystyle
I_{3a}=\int_2^4 \frac{dx}{x( ln\, x)^2}=\ln\left({2}\right) \\
u=\ln\left({x}\right) \ \ \ du=\frac{1}{x}\, dx $$
$\text{then} $
$$\displaystyle
I_{3a}
=\int_2^4\frac{1}{{u}^{2}}\, du
= \left[-\frac{1}{u}\right]_2^4$$
So far??

The only thing you neglected in this approach was to change the limits...

$$I=\int_{\ln(2)}^{\ln(4)}\frac{1}{u^2}\, du=-\left[\frac{1}{u}\right]_{\ln(2)}^{\ln(4)}=\frac{1}{\ln(2)}-\frac{1}{\ln(4)}=\frac{1}{2\ln(2)}$$

 

[karush]

Re: 242.q2.3a Int_2^4 dx/(x ln x)^2 dx

The only thing you neglected in this approach was to change the limits...

$$I=\int_{\ln(2)}^{\ln(4)}\frac{1}{u^2}\, du=-\left[\frac{1}{u}\right]_{\ln(2)}^{\ln(4)}=\frac{1}{\ln(2)}-\frac{1}{\ln(4)}=\frac{1}{2\ln(2)}$$

 

[LavonBuchf]

Re: 242.q2.3a Int_2^4 dx/(x ln x)^2 dx

That was quite tricky, but solved.

 

[HallsofIvy]

Re: 242.q2.3a Int_2^4 dx/(x ln x)^2 dx

I notice that the integral in the title of this thread is \int_2^4\frac{1}{(x ln(x))^2}dx[/int] which is quite a different problem!

 

[MarkFL]

Re: 242.q2.3a Int_2^4 dx/(x ln x)^2 dx

I notice that the integral in the title of this thread is $$\int_2^4\frac{1}{(x ln(x))^2}dx$$ which is quite a different problem!

Thanks for the catch!...I have edited the thread title accordingly. :D