Integration of (3x+1) / (2x^2 - 2x +3 )

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In summary, the result of the integral will contain a ln function and an arctan function. If it does have real solutions, you can try partial fraction.
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teng125
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integration of (3x+1) / (2x^2 - 2x +3 )
pls help...thank you
 
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A u-subtitution would probably work well.
 
  • #3
?
You have posted some problem that's similar to this before. You can click here.
Please read the posts there and think thoroughly (think about the way to solve it, why people are using that way, is there any quicker way, etc...), as there will be no-one who will be able to give you a hint when you are in an exam!
The result of the integral
[tex]\int \frac{ax + b}{g x ^ 2 + h x + i} dx[/tex] (a, b, c, g, h, i are all constants), will contain a ln function and an arctan function. (Note that: it does not contain an arctan part if the numerator is the derivative of the denominator) (if the denominator does not have real solutions).
If it does have real solutions, you can try partial fraction. There's a thread about partial fraction in the tutorial board.
Example of the solution contains no arctan part:
[tex]\int \frac{2x + 1}{x ^ 2 + x + 3} dx = \ln |x ^ 2 + x + 3| + C[/tex].
The numerator is the derivative of the denominator, it can be dome with a u-substitution: u = denominator (in this example: u = x2 + x + 3).
Take the derivatives of the denominator, ie: 2gx + h.
Split the numerator into 2 parts, 1 part is of the form: [itex]\alpha(2gx + h)[/itex], the other part is a constant.
From there, you will have:
[tex]\int \frac{ax + b}{g x ^ 2 + h x + i} dx = \int \frac{a}{2g} \ \frac{2gx + b\frac{2g}{a}}{g x ^ 2 + h x + i} dx = \frac{a}{2g} \int \frac{2gx + h + b\frac{2g}{a} - h}{g x ^ 2 + h x + i} dx[/tex]
[tex]= \frac{a}{2g} \int \frac{2gx + h}{g x ^ 2 + h x + i} dx + \int \frac{b - h \ \frac{a}{2g}}{gx ^ 2 + h x + i} dx[/tex]
You will have 2 separate integrals, the first one can be done with a u-substitution u = gx2 + hx + i, the other will give you an arctan part (try completing the square in the denominator).
-----------------------
Example:
[tex]\int \frac{3x + 5}{x ^ 2 + x + 2} dx[/tex]
x2 + x + 2 does not have real solutions. The derivative of it is 2x + 1.
[tex]\int \frac{3x + 5}{x ^ 2 + x + 2} dx = \frac{3}{2} \int \frac{2x + \frac{10}{3}}{x ^ 2 + x + 2} dx = \frac{3}{2} \int \frac{(2x + 1) + \frac{7}{3}}{x ^ 2 + x + 2} dx = \frac{3}{2} \ln|x ^ 2 + x + 1| + \frac{7}{2} \int \frac{dx}{x ^ 2 + x + 2}[/tex]
[tex]= \frac{3}{2} \ln|x ^ 2 + x + 2| + \frac{7}{2} \int \frac{dx}{ \left( x + \frac{1}{2} \right) ^ 2 + \frac{7}{4}} = \frac{3}{2} \ln|x ^ 2 + x + 2| + \frac{7}{2} \ \frac{2}{\sqrt 7} \arctan \left( 2 \ \frac{x + \frac{1}{2}}{\sqrt{7}} \right) + C[/tex]
[tex]= \frac{3}{2} \ln|x ^ 2 + x + 2| + \sqrt 7 \arctan \left(\frac{2x + 1}{\sqrt{7}} \right) + C[/tex].
Now read the example, think, and try to do your problem again. Just shout out if you get stuck somewhere.
 
Last edited:

1. What is the formula for integrating the expression (3x+1) / (2x^2 - 2x +3)?

The formula for integrating the expression (3x+1) / (2x^2 - 2x +3) is ∫(3x+1) / (2x^2 - 2x +3) dx = (3/4)ln|2x^2 - 2x +3| + C.

2. How do you solve the integral of (3x+1) / (2x^2 - 2x +3)?

To solve the integral of (3x+1) / (2x^2 - 2x +3), you can use the formula ∫(3x+1) / (2x^2 - 2x +3) dx = (3/4)ln|2x^2 - 2x +3| + C. You can also use techniques such as u-substitution or partial fractions to simplify the integral before integrating.

3. Can the expression (3x+1) / (2x^2 - 2x +3) be integrated using the power rule?

No, the expression (3x+1) / (2x^2 - 2x +3) cannot be integrated using the power rule. It requires more advanced techniques such as u-substitution or partial fractions.

4. What is the domain of the expression (3x+1) / (2x^2 - 2x +3)?

The domain of the expression (3x+1) / (2x^2 - 2x +3) is all real numbers except for x = 1. This is because when x = 1, the denominator becomes 0, which is undefined.

5. How does the graph of (3x+1) / (2x^2 - 2x +3) look like?

The graph of (3x+1) / (2x^2 - 2x +3) is a hyperbola that opens upwards and has a horizontal asymptote at y = 0. It also has a vertical asymptote at x = 1. The graph approaches the x-axis but never touches it, and it approaches the vertical asymptote but never crosses it.

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