Evaluate Integral: 2 ln| \frac{v-1}{v}|

In summary: It's easy to forget that the variable of integration (in this case, v) is not the same as the variable in the denominator (in this case, v^2-v). So, you cannot simply replace the v in the denominator with x and use the rule for integrating 1/x. You must follow the appropriate steps for integrating fractions with more complicated denominators, as others have shown you. Keep practicing and it will become second nature to you.
  • #1
whatlifeforme
219
0

Homework Statement


evaluate the integral.


Homework Equations


[itex]\displaystyle\int_2^∞ {\frac{2}{v^2 -v} dv}[/itex]


The Attempt at a Solution


how does this integrate into:

2 ln| [itex]\frac{v-1}{v}| [/itex]

i tried and got 2ln|v^2-v| but not above.
 
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  • #2
whatlifeforme said:

Homework Statement


evaluate the integral.


Homework Equations


[itex]\displaystyle\int_2^∞ {\frac{2}{v^2 -v} dv}[/itex]


The Attempt at a Solution


how does this integrate into:

2 ln| [itex]\frac{v-1}{v}| [/itex]

i tried and got 2ln|v^2-v| but not above.

Did you try factoring and partial fractions?
 
  • #3
Try simplifying the denominator and using the method of partial fractions. Ahhh beaten to the punch!
 
  • #4
Also since you know (knew?) the answer, differentiate it and you see what is right and may get some helpful insight and reinforcement.
 
  • #5
whatlifeforme said:
[itex]\displaystyle\int_2^∞ {\frac{2}{v^2 -v} dv}[/itex]

The Attempt at a Solution


how does this integrate into:

2 ln| [itex]\frac{v-1}{v}| [/itex]

i tried and got 2ln|v^2-v| but not above.

Others have shown you the right way - I'll explain what you did that was wrong.

These are correct:
$$\int \frac{dx}{x} = ln|x| + C$$
$$\int \frac{du}{u} = ln|u| + C~$$

BUT, this is NOT correct:
$$\int \frac{dx}{f(x)} = ln|f(x)| + C$$

This is a very common error among students who are learning calculus.
 

Related to Evaluate Integral: 2 ln| \frac{v-1}{v}|

1. What is the formula for evaluating this integral?

The formula for evaluating the integral of 2 ln| \frac{v-1}{v}| is \int 2 \ln \left| \frac{v-1}{v} \right| dv = 2v \ln \left| \frac{v-1}{v} \right| - 2v + C.

2. What is the domain of this integral?

The domain of this integral is all real numbers except for v = 0 and v = 1.

3. How do you handle the absolute value in the integrand?

To handle the absolute value, you can split the integral into two parts: \int 2 \ln \left( \frac{v-1}{v} \right) dv + \int 2 \ln \left( \frac{1-v}{v} \right) dv. Then, you can use the properties of logarithms to simplify the integrals.

4. Can this integral be solved using substitution?

Yes, this integral can be solved using substitution. A possible substitution is u = v-1, which leads to du = dv. This substitution will help simplify the integral.

5. Are there any special techniques or strategies for solving this integral?

One helpful strategy for solving this integral is to use integration by parts. This can be done by setting u = ln| \frac{v-1}{v}| and dv = 2dv. Then, you can use the formula \int udv = uv - \int vdu to simplify the integral.

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