Integration by parts help natural log hurry tired

In summary, the conversation discusses how to integrate the function (3x)/(3x-2) using integration by parts or substitution. The speakers also mention a trick to avoid using integration by parts and discuss simplifying the expression by polynomial long division. There is also confusion about whether the answer should involve absolute values or not.
  • #1
darthxepher
56
0
Integrate the function. (3x)/(3x-2)

I am spose to use integration by parts but i don't know how to integrate 1/(3x-2). anybody help??
 
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  • #2
[tex]\int \frac{dx}{3x-2}[/tex] well, i don't know why ure supposed to use integ. by parts, since a nice subst. would work.

substitute 3x-2=u, then 3dx=du=> dx=du/3 so

[tex]\int \frac{dx}{3x-2}=\frac{1}{3}\int \frac{du}{u}=\frac{1}{3}ln|u|+c[/tex] now just go back to the original variable x.
 
  • #3
Well, i assume now that you were asked to use integ. by parts on the original function. but if you are not required to do so, there is a nice trick to avoid it ,so you will only need to use subst.
 
  • #4
I tried but it said the answer doesn't involve absolute value... :( the original problem was find the integral of >>> ln(3x-2)
 
  • #5
well, if you drop the absolute values, then you are assuming that 3x-2>0, otherwise that would not hold.
 
  • #6
But wait i asked to integrate (3x)/(3x-2) not dx/(3x-2)... hm... I am confused
 
  • #7
darthxepher said:
I tried but it said the answer doesn't involve absolute value... :( the original problem was find the integral of >>> ln(3x-2)
You should have told us so explicitly. Use integration by parts here, denote u=ln(3x-2). Find v and du, and then you'll have to integrate 3x/(3x-2) which you should do by first simplifying the expression by polynomial long division. And are you sure the answer doesn't have absolute signs? There is a ln in the answer is there not?
 
Last edited:
  • #8
darthxepher said:
But wait i asked to integrate (3x)/(3x-2) not dx/(3x-2)... hm... I am confused
Yeah, but look here

[tex] \frac{3x}{3x-2}=\frac{3x-2+2}{3x-2}=\frac{3x-2}{3x-2}+\frac{2}{3x-2}=1+\frac{2}{3x-2}[/tex]

so all you need to integrate is what i already told you how!...lol...
 

Related to Integration by parts help natural log hurry tired

1. What is integration by parts?

Integration by parts is a mathematical technique used to evaluate integrals that are products of two functions. It involves breaking down the integral into smaller parts and using the product rule of differentiation to solve it.

2. How does integration by parts work?

Integration by parts works by using the product rule of differentiation to simplify the integral into a more manageable form. The formula for integration by parts is ∫u dv = uv - ∫v du, where u and v are two functions and du and dv are their respective differentials.

3. What is the natural log function?

The natural log function, denoted as ln(x), is the inverse of the exponential function. It is a logarithmic function with a base of e, where e is a mathematical constant approximately equal to 2.718. The natural log function is commonly used in mathematics, especially in calculus and logarithmic differentiation.

4. How can integration by parts help with solving natural log integrals?

Integration by parts can help solve natural log integrals by breaking down the integral into two parts and using the product rule of differentiation. This allows for the integral to be simplified and solved using known integration techniques.

5. Why is it important to use integration by parts when dealing with natural log integrals?

It is important to use integration by parts when dealing with natural log integrals because they can be difficult to solve using other integration techniques. Integration by parts provides a systematic way to simplify the integral and make it easier to solve. It is also useful when dealing with more complex integrals that involve natural log and other functions.

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