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karush
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$\Large{242.Q2.3a} \\
\text{find the integral} \\
\displaystyle
\int_2^4 \frac{dx}{x( ln\, x)^2} $
\text{find the integral} \\
\displaystyle
\int_2^4 \frac{dx}{x( ln\, x)^2} $
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One problem is that you have too many "dx"s. :)karush said:$\Large{242.Q2.3a} \\
\text{find the integral} \\
\displaystyle
\int_2^4 \frac{dx}{x( ln\, x)^2} dx$
Yes this has 2 dx so ??
karush said:$\displaystyle \text{Find the integral: } \; \int_2^4 \frac{dx}{x(\ ln x)^2} $
You didn't back substitute in you last step. It's \(\displaystyle \left[-\frac{1}{\ln{x}}\right]karush said:$\Large{242.Q2.3a} \\
\text{find the integral} $
$$\displaystyle
I_{3a}=\int_2^4 \frac{dx}{x( ln\, x)^2}=\ln\left({2}\right) \\
u=\ln\left({x}\right) \ \ \ du=\frac{1}{x}\, dx $$
$\text{integrate and back substitute } u=ln(x)$
$$\displaystyle
I_{3a}
=\int_{\ln{2}}^{2\ln{2}}\frac{1}{{u}^{2}}\, du
= \left[-\frac{1}{\ln{x}}\right]
_{\ln\left({2}\right)}^{2\ln\left({2}\right)}$$
karush said:$\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 said:The only thing you neglected in this approach was to change the limits...
\(\displaystyle I=\int_{\ln(2)}^{\ln(4)}\frac{1}{u^2}\, du=-\left[\frac{1}{u}\right]_{\ln(2)}^{\ln(4)}=\frac{1}{\ln(2)}-\frac{1}{\ln(4)}=\frac{1}{2\ln(2)}\)
HallsofIvy said:I notice that the integral in the title of this thread is \(\displaystyle \int_2^4\frac{1}{(x ln(x))^2}dx\) which is quite a different problem!
The expression "242.q2.3a Int_2^4 dx/[x( ln x)^2]" represents a definite integral, a mathematical concept used to find the area under a curve within a specific interval. In this case, it is the area under the curve of the function 1/[x( ln x)^2] between the values of 2 and 4.
This integral can be solved using various techniques, such as substitution, integration by parts, or partial fractions. The specific method used will depend on the complexity of the integrand (the function being integrated).
The denominator of the integrand, [x( ln x)^2], is squared because it is an indeterminate form. This means that as x approaches 0, the function approaches infinity. Squaring the denominator helps to eliminate this indeterminacy and allows us to find the definite integral.
The interval [2,4] is the bounds of integration, which specify the range of values for x over which the integral is being evaluated. In this case, the integral is being evaluated from x = 2 to x = 4, meaning we are finding the area under the curve of the function between these two values.
Integrals have many applications in various fields of science, including physics, engineering, and economics. In this specific integral, the function 1/[x( ln x)^2] can be used to model certain physical phenomena, such as the decay of radioactive materials. The integral can also be used to find the average value of the function over the given interval or to calculate probabilities in certain statistical distributions.