Solve the given problem involving integration

[chwala]

Homework Statement

see attached

Relevant Equations

integration

This really cracked me up! Unless there is something i am not seeing!

part (a) is straightforward, using quotient rule:
$\dfrac{dy}{dx} = \dfrac{x⋅\dfrac{1}{x}- \ln x}{x^2}=\dfrac{1-\ln x}{x^2}$

From here i was able to see that,

$\int \dfrac{\ln x}{x^2} dx= \int \dfrac{1}{x^2}- \dfrac{\ln x}{x}$
$= -\dfrac{1}{x} - \dfrac{ln x}{x}=-\left[\dfrac{1+\ln x}{x}\right]$

on the other hand, using integration by parts( without using part (a) and (b)), gives me
$\int \dfrac{\ln x}{x^3} dx =\dfrac{-\ln x}{2x^2} + \int \left[\dfrac{-1}{2x^2}⋅\dfrac{1}{x}\right] dx$

$= -\dfrac{\ln x}{2x^2}+ \int \dfrac{-1}{2x^3} dx$

$= -\dfrac{\ln x}{2x^2}-\dfrac{1}{4x^2} +c$

but this does not satisfy my curiosity, as i did not use part (a), was there a mistake on the textbook or am i simply missing out on something?

 

[PhDeezNutz]

So we have as you said

Equation 1
$\frac{d}{dx}\left(\frac{ \ln x}{x} \right) = \frac{1- \ln x}{x^2} = \frac{1}{x^2} - \frac{\ln x}{x^2}$

Which implies

Equation 2
$\frac{\ln x}{x^2} = \frac{1}{x^2} - \frac{d}{dx}\left(\frac{ \ln x}{x} \right)$ This becomes important later

Starting with the integral

$\int \frac{\ln x} {x^3} \, dx = \int \frac{1}{x} \left(\frac{\ln x} {x^2} \right)\, dx$

substitute Equation 2 into the above

$\int \frac{\ln x} {x^3} \, dx = \int \frac{1}{x} \left( \frac{1}{x^2} - \frac{d}{dx}\left(\frac{ \ln x}{x} \right)\right)\,dx$

$\int \frac{\ln x} {x^3} \, dx = \int \left(\frac{1}{x^3} - \frac{1}{x} \frac{d}{dx} \left( \frac{\ln x}{x}\right)\right)\, dx$
Well the first term on the right hand side of the equals sign is easy enough to integrate readily

$\int \frac{\ln x} {x^3} \, dx = - \frac{1}{2x^2} - \int \frac{1}{x} \frac{d}{dx} \left( \frac{\ln x}{x} \right)\,dx$

Now use integration by parts on the integral on the right hand side of the equals sign

$v = \frac{\ln x}{x}$
$dv = \frac{d}{dx} \left(\frac{\ln x}{x} \right) dx$
$u = \frac{1}{x}$
$du = - \frac{1}{x^2}$

Plugging this all in

$\int \frac{\ln x}{x^3} \,dx = -\frac{1}{2x^2} - \frac{\ln x}{x^2} - \int \frac{\ln x}{x^3} \,dx$

$2 \int \frac{\ln x}{x^3} \,dx = -\frac{1}{2x^2} - \frac{\ln x}{x^2}$

$\int \frac{\ln x}{x^3} \,dx = -\frac{1}{4x^2} - \frac{\ln x}{2x^2} + C$

I get the same thing as you

 

[fresh_42]

... but this does not satisfy my curiosity, as i did not use part (a), was there a mistake on the textbook or am i simply missing out on something?

Yes. Firstly, you made a sign error. Twice, so that your result remained correct. Secondly, as @PhDeezNutz demonstrated, you missed a trick by not using part (a) of the problem. You solved it by integration by parts
$$
\int (u'v)=uv-\int (uv')
$$
which is faster and more suitable in this case. However, using part (a) would have meant to use
$$
\mathcal{I}:=\int \dfrac{\log(x)}{x^3}=-\int \left(\underbrace{\dfrac{1-\log(x)}{x^2}}_{u'}\cdot \underbrace{\dfrac{1}{x}}_{v}-\dfrac{1}{x^2}\right)
$$
which results in an equation
$$
\mathcal{I}=f(x) -\mathcal{I} \Longleftrightarrow \mathcal{I}=\dfrac{f(x)}{2}
$$

The trick to using the additive symmetry: an integral that shows up on both sides of the equation with an opposite sign. You do not "solve" the integral, you express it with another function: $f(x).$ A similar trick exists for multiplicative symmetries.

 

[fresh_42]

In case you, @chwala, or other readers are interested in a more complicated example you can look at
https://www.physicsforums.com/threads/solution-manuals-for-the-math-challenges.977057/
(June 2020, ex. 2, proof of the hint) and (September 2019, ex. 14) for an example of multiplicative symmetry.

 

[anuttarasammyak]

From a)
\frac{d^2}{dx^2}\frac{\ln x }{x}=-\frac{3}{x^3}+\frac{2}{x^3}\ln x
\frac{\ln x}{x^3} = \frac{1}{2}\frac{d^2}{dx^2}\frac{\ln x }{x}+\frac{3}{2x^3}
\int dx\frac{\ln x}{x^3} = \frac{1}{2}\frac{d}{dx}\frac{\ln x }{x}+\int dx\frac{3}{2x^3}
So the same result with you @chwala .

 

[PhDeezNutz]

From a)
\frac{d^2}{dx^2}\frac{\ln x }{x}=-\frac{3}{x^3}+\frac{2}{x^3}\ln x
\frac{\ln x}{x^3} = \frac{1}{2}\frac{d^2}{dx^2}\frac{\ln x }{x}+\frac{3}{2x^3}
\int dx\frac{\ln x}{x^3} = \frac{1}{2}\frac{d}{dx}\frac{\ln x }{x}+\int dx\frac{3}{2x^3}
So the same result with you @chwala .

I really liked how you used the second derivative. Even though that’s not what the problem asked for I can still definitely appreciate it.

Edit: nvm you only used the second derivative so you could integrate it and get the first derivative and use part a) which is clever. Either way pretty cool.

 

[chwala]

Yes. Firstly, you made a sign error. Twice, so that your result remained correct. Secondly, as @PhDeezNutz demonstrated, you missed a trick by not using part (a) of the problem. You solved it by integration by parts
$$
\int (u'v)=uv-\int (uv')
$$
which is faster and more suitable in this case. However, using part (a) would have meant to use
$$
\mathcal{I}:=\int \dfrac{\log(x)}{x^3}=-\int \left(\underbrace{\dfrac{1-\log(x)}{x^2}}_{u'}\cdot \underbrace{\dfrac{1}{x}}_{v}-\dfrac{1}{x^2}\right)
$$
which results in an equation
$$
\mathcal{I}=f(x) -\mathcal{I} \Longleftrightarrow \mathcal{I}=\dfrac{f(x)}{2}
$$

The trick to using the additive symmetry: an integral that shows up on both sides of the equation with an opposite sign. You do not "solve" the integral, you express it with another function: $f(x).$ A similar trick exists for multiplicative symmetries.

I think it is fair that i fix the sign error, i will proceed as follows,

$$\int \dfrac{\ln x}{x^3} dx =\dfrac{-\ln x}{2x^2} - \int \left[\dfrac{-1}{2x^2}⋅\dfrac{1}{x}\right] dx + c$$

$$ =\dfrac{-\ln x}{2x^2} + \int \left[\dfrac{1}{2x^3}\right] dx+c$$

$$=\dfrac{-\ln x}{2x^2} +\dfrac{1}{2} \int \left[\dfrac{1}{x^3}\right] dx+c$$

$$=\dfrac{-\ln x}{2x^2} +\dfrac{1}{2} ⋅ \left(\dfrac{-1}{2x^2}\right)+c$$

$$= -\dfrac{\ln x}{2x^2}-\dfrac{1}{4x^2} +c$$