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

1. Jan 19, 2006

### teng125

integration of (3x+1) / (2x^2 - 2x +3 )
pls help.....thank you

2. Jan 19, 2006

### Mindscrape

A u-subtitution would probably work well.

3. Jan 20, 2006

### VietDao29

???
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
$$\int \frac{ax + b}{g x ^ 2 + h x + i} dx$$ (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:
$$\int \frac{2x + 1}{x ^ 2 + x + 3} dx = \ln |x ^ 2 + x + 3| + C$$.
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: $\alpha(2gx + h)$, the other part is a constant.
From there, you will have:
$$\int \frac{ax + b}{g x ^ 2 + h x + i} dx = \int \frac{a}{2g} \ \frac{2gx + b\frac{2g}{a}}{g x ^ 2 + h x + i} dx = \frac{a}{2g} \int \frac{2gx + h + b\frac{2g}{a} - h}{g x ^ 2 + h x + i} dx$$
$$= \frac{a}{2g} \int \frac{2gx + h}{g x ^ 2 + h x + i} dx + \int \frac{b - h \ \frac{a}{2g}}{gx ^ 2 + h x + i} dx$$
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).
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Example:
$$\int \frac{3x + 5}{x ^ 2 + x + 2} dx$$
x2 + x + 2 does not have real solutions. The derivative of it is 2x + 1.
$$\int \frac{3x + 5}{x ^ 2 + x + 2} dx = \frac{3}{2} \int \frac{2x + \frac{10}{3}}{x ^ 2 + x + 2} dx = \frac{3}{2} \int \frac{(2x + 1) + \frac{7}{3}}{x ^ 2 + x + 2} dx = \frac{3}{2} \ln|x ^ 2 + x + 1| + \frac{7}{2} \int \frac{dx}{x ^ 2 + x + 2}$$
$$= \frac{3}{2} \ln|x ^ 2 + x + 2| + \frac{7}{2} \int \frac{dx}{ \left( x + \frac{1}{2} \right) ^ 2 + \frac{7}{4}} = \frac{3}{2} \ln|x ^ 2 + x + 2| + \frac{7}{2} \ \frac{2}{\sqrt 7} \arctan \left( 2 \ \frac{x + \frac{1}{2}}{\sqrt{7}} \right) + C$$
$$= \frac{3}{2} \ln|x ^ 2 + x + 2| + \sqrt 7 \arctan \left(\frac{2x + 1}{\sqrt{7}} \right) + C$$.
Now read the example, think, and try to do your problem again. Just shout out if you get stuck somewhere.

Last edited: Jan 20, 2006