Integral of (x-4)/(x^2-x+2)

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  • #1
Nicolaus
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Homework Statement


[itex]\frac{x-4}{x^2-x+2}[/itex]

Homework Equations


The Attempt at a Solution



[itex]\frac{x-4}{(x-1/2)^2+7/4}[/itex]
u = x - 1/2 = [itex]\frac{\sqrt{7}}{2}[/itex]tan[itex]\theta[/itex] ; du = [itex]\sqrt{7}[/itex]/2 sec^2(theta)
x = u + 1/2
[itex]\int\frac{(u−7/2)(\sqrt{7}/2sec^2(\theta)}{7/4 sec^2(\theta)}[/itex]

This simplifies to: [itex]\int tan(\theta) - \sqrt{7}[/itex]
= ln(sec(theta)) - [itex]\sqrt{7}[/itex](theta)
= ln([itex]\frac{\sqrt{(2x-1)^2+7}}{\sqrt{7}}[/itex]) - [itex]\sqrt{7}[/itex]arctan([itex]\frac{2x-1}{\sqrt{7}}[/itex]) + C
 
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  • #2
Nicolaus said:

Homework Statement


[itex]\frac{x-4}{x^2-x+2}[/itex]


Homework Equations





The Attempt at a Solution



[itex]\frac{x-4}{(x-1/2)^2+7/4}[/itex]
u = x - 1/2 = [itex]\frac{\sqrt{7}}{2}[/itex]tan[itex]\theta[/itex] ; du = [itex]\sqrt{7}[/itex]/2 sec^2(theta)
x = u + 1/2
[itex]\int\frac{(u−7/2)(\sqrt{7}/2sec^2(\theta)}{7/4 sec^2(\theta)}[/itex]

This simplifies to: [itex]\int tan(\theta) - \sqrt{7}[/itex]
= ln(sec(theta)) - [itex]\sqrt{7}[/itex](theta)
= ln([itex]\frac{\sqrt{(2x-1)^2+7}}{\sqrt{7}}[/itex]) - [itex]\sqrt{7}[/itex]arctan([itex]\frac{2x-1}{\sqrt{7}}[/itex]) + C

Do you have a question?

You can check your answer by differentiating. If you end up with the same as the original integrand, your answer is correct. It's pretty messy, but if you use log properties before you differentiate, that will make things a lot simpler.

Another approach that is probably simpler is to split up the integrand using partial fractions.
 
  • #3
I can't think of any "tricks" to solving this. I don't see how you could use partial fractions either... Did you make this problem up?
 
  • #4
Yeah, on second thought, partial fractions wouldn't be as easy as I thought. The factors would include complex numbers.
 
  • #5
It is quite easy to solve the given integral.

Rewrite it as
$$\frac{1}{2}\int \frac{2x-1}{x^2-x+2}dx-\frac{7}{2}\int\frac{dx}{x^2-x+2}$$

Both the integrals can be easily solved.
 
  • #6
Pranav-Arora said:
It is quite easy to solve the given integral.

Rewrite it as
$$\frac{1}{2}\int \frac{2x-1}{x^2-x+2}dx-\frac{7}{2}\int\frac{dx}{x^2-x+2}$$

Both the integrals can be easily solved.

How did you even think of that o.o and it still is pretty difficult to solve the 2nd one.
 
  • #7
iRaid said:
How did you even think of that o.o and it still is pretty difficult to solve the 2nd one.

You complete the square, just like Nicolaus did in the OP. Everybody here knows that the answer in the first post is correct, yes? The only problem with it is that the log term is in a pretty awkward form.
 
  • #8
I am just wondering if that is correct. Ya, the log form is messy.
To simplify that:

ln((2x−1)2+7√7√) - 7√arctan(2x−17√) + C
= 1/2(ln(2x-1)^2+1) - ln(7))

So, the integral in the OP is correct, right?
 
  • #9
Nicolaus said:
I am just wondering if that is correct. Ya, the log form is messy.
To simplify that:

ln((2x−1)2+7√7√) - 7√arctan(2x−17√) + C
= 1/2(ln(2x-1)^2+1) - ln(7))

So, the integral in the OP is correct, right?

Yes it's correct, but I could read the integral in the OP. I can't read what you have written there or what you are thinking.
 
  • #10
The solution looks a bit simpler in the form ##1/2 \ln |x^2-x+2|-\sqrt7 \arctan(\frac{2x-1}{\sqrt7})+C##

ehild
 
  • #11
iRaid, the question can not be solved by partial fractions. The method is as follows :

You have to get the form Pranav got. I would tell you how ?

Method I: By inspection: Just break the fraction and try to get the form f'(x)dx/f(x)...

Method II :

Let,

x-4=λd(x2-x+2)/dx + μ

You can find λ, and μ by comparing the coefficients. Then break the fraction to get,

FORM : Integral of Derivative of function/Function + Integration of reciprocal of a quadratic.

The second one is integrated by method of perfect squares which I hope you know.
 
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1. What is the formula for the integral of (x-4)/(x^2-x+2)?

The formula for the integral of (x-4)/(x^2-x+2) is ∫(x-4)/(x^2-x+2)dx = ln|x^2-x+2| + C, where C is the constant of integration.

2. How do you solve the integral of (x-4)/(x^2-x+2)?

To solve the integral of (x-4)/(x^2-x+2), you can use the substitution method by letting u = x^2-x+2. Then, du = (2x-1)dx and the integral becomes ∫(x-4)/(x^2-x+2)dx = ∫1/u du = ln|u| + C = ln|x^2-x+2| + C.

3. Can the integral of (x-4)/(x^2-x+2) be solved using partial fractions?

Yes, the integral of (x-4)/(x^2-x+2) can be solved using partial fractions. However, the resulting partial fractions may not be easily integrable and the substitution method may be a better approach.

4. What is the domain of the integral of (x-4)/(x^2-x+2)?

The domain of the integral of (x-4)/(x^2-x+2) is all real numbers except for x = 1 and x = -2, where the denominator becomes 0.

5. Is there a graphical interpretation of the integral of (x-4)/(x^2-x+2)?

Yes, the integral of (x-4)/(x^2-x+2) can be interpreted as the area under the curve of the function f(x) = (x-4)/(x^2-x+2) from x = a to x = b. This area can be approximated using Riemann sums and can be visualized on a graph.

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