Integrating 1/xln(x) using integration by parts

In summary: The key to understanding why this happens is to remember that when you integrate by parts, you're actually solving for the indefinite integral, not the definite integral. So when you solve for I using integration by parts, you end up with a new indefinite integral, which still has a constant of integration. That's why when you plug I back into the original equation, you get 1 = C, where C is just a different constant than the one you started with. Therefore, when you solve for I using integration by parts and plug it back in, you're not actually solving the same integral as when you used substitution. The two methods yield different constants, hence the different solutions. In summary, the two methods will yield different solutions because they are solving
  • #1
ductruffe
3
0

Homework Statement


Integrating 1/xlnx by parts...

Homework Equations


Find the integral of 1/xlnx

The question asks to solve by substitution, which I can do and results in ln(ln(x)) + c

It then asks to compute using integration by parts, and then to explain how it can be true (because it will compute something different to substitution).

The Attempt at a Solution


I = uv - int (v dU)

let u= 1/lnx du = 1/x(lnx)^2
let dv = 1/x, v = lnx

Sub into the parts formula

I = lnx* 1/lnx - int (lnx/x(lnx)^2)
I = lnx/lnx - int (1/xlnx) <--- what we started with
I = 1 - int (1/xlnx) This is 1 - I, the integral we began with...

I've bene shown this trick where you can go..
I = 1- I
2I = 1
I = 1/2

I'm not sure if this is correct, but I would appreciate any help

Thank you
 
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  • #2
It is correct
 
  • #3
Thank you for the quick reply. I'm also asked to explain how I = 1/2 can be true, when using substitution yields ln(ln(x)). This is basically where I am stuck.

Thank you
 
  • #4
ln(ln(x)) is correct.
Regarding Integration by parts, you have missed -ve sign in differentiating u
It will be
I=1+I
resulting in 1=0 which is wrong.
 
  • #5
n.karthick said:
ln(ln(x)) is correct.
Regarding Integration by parts, you have missed -ve sign in differentiating u
It will be
I=1+I
resulting in 1=0 which is wrong.

No, it results in 1 = C.
 
  • #6
I can see I've missed the negative which changes it quite a bit.

Is the integration by parts correct for C=1? It hasn't really solved the integral, or am I missing something here?
 
  • #7
Typically, you'll see this problem as an example of why it's so important to remember the constant of integration, because otherwise you end up with nonsense like 1 = 0.
 

1. What is integration by parts?

Integration by parts is a method of integration used to find the integral of a product of two functions. It involves breaking down the integral into two parts and using a formula to rewrite the integral in a different form.

2. How do you integrate 1/xln(x) using integration by parts?

To integrate 1/xln(x) using integration by parts, you would first choose one part of the integral to be the "u" term and the other part to be the "dv" term. Then, you would use the formula: ∫udv = uv - ∫vdu, where "u" and "v" refer to the chosen terms, to rewrite the integral. This process is repeated until the integral can be easily evaluated.

3. What is the formula for integration by parts?

The formula for integration by parts is: ∫udv = uv - ∫vdu. This formula is used to rewrite an integral involving a product of two functions into a form that is easier to integrate.

4. When should I use integration by parts?

Integration by parts is useful when the integral involves a product of two functions and other integration techniques, such as substitution, are not applicable. It is also helpful when one of the functions in the integral can be easily differentiated and the other can be easily integrated.

5. Are there any limitations to using integration by parts?

Yes, there are limitations to using integration by parts. This method can only be used for certain types of integrals, specifically those involving a product of two functions. It may not always be the most efficient or effective method of integration, and in some cases, other techniques may be more suitable.

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