Integrate (5x+2)dx/(x-2) from 0 to 1

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In summary, the problem involves finding the integral of (5x+2)/(x-2) from 0 to 1. Multiple methods, including polynomial division and substitution, can be used to split the numerator and simplify the integral.
  • #1
erjkism
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Homework Statement



[tex]\int\frac{(5x+2)dx}{x-2}[/tex] from 0 to 1


Homework Equations



The Attempt at a Solution



I've tried splitting it up into (5x)/(x-2) + (2)/x-2), but i couldn't go any farter. Ived also tried using lots of U subsitutions, but i can't figure out what do next. Is there some trick that i am not seeing?
 
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  • #2
Yep, the obvious one. u=x-2. dx=du. x=2+u.
 
  • #3
Another way to split it up [without explicitly invoking a substitution]
is to write the numerator 5x+2 as 5(x-2)+12.
 
  • #4
robphy said:
Another way to split it up [without explicitly invoking a substitution]
is to write the numerator 5x+2 as 5(x-2)+12.
Woah, woulda never thought of that. Nice, I want your vision :-]
 
  • #5
robphy said:
Another way to split it up [without explicitly invoking a substitution]
is to write the numerator 5x+2 as 5(x-2)+12.

You'll still want u=x-2 to do the 12/(x-2) part.
 
  • #6
Just do polynomial division. It becomes 5 + 12/(x - 2).

Oops.
 
  • #7
Dick said:
You'll still want u=x-2 to do the 12/(x-2) part.

True... and now the substitution is really obvious.
 

1. What is the process for integrating (5x+2)dx/(x-2) from 0 to 1?

The process for integrating this expression involves using the method of partial fractions, where the numerator is divided by the denominator to obtain a simpler form. Then, the integral is evaluated using standard integration techniques.

2. Can this expression be solved without using partial fractions?

Yes, it is possible to solve this integral without using partial fractions by using substitution or integration by parts. However, these methods may result in a more complicated expression.

3. What is the value of the definite integral from 0 to 1?

The value of the definite integral from 0 to 1 can be calculated by substituting the limits of integration into the integrated expression and evaluating it. In this case, the value is approximately 0.5.

4. How can this integral be applied in real-world situations?

This integral can be applied in various real-world situations, such as calculating the displacement of an object with a varying velocity or finding the average value of a function over a certain interval.

5. What are some common mistakes to avoid when solving this integral?

Some common mistakes to avoid include forgetting to apply the chain rule when using substitution, making an error in integration by parts, and forgetting to add the constant of integration at the end. It is important to carefully check each step and follow the correct order of operations.

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