# Homework Help: Integral of a fraction consisting of two quadratic equations

1. Jul 5, 2008

### phyzmatix

1. The problem statement, all variables and given/known data
Determine

$$\int\frac{4x^{2}-2x+2}{4x^{2}-4x+3}dx$$

2. Relevant equations
I believe it's necessary to complete the square.

3. The attempt at a solution
Completing the square for

$$4x^{2}-4x+3$$

gives

$$\int\frac{4x^{2}-2x+2}{(x-\frac{1}{2})^2+\frac{1}{2}}dx$$

let

$$u=x-\frac{1}{2}, du = dx, x=u+\frac{1}{2}$$

then

$$\int\frac{4(u+\frac{1}{2})^{2}-2(u+\frac{1}{2})+2}{u^2+\frac{1}{2}}du$$
$$=\int\frac{4u^2+4u+1-2u-1+2}{u^2+\frac{1}{2}}du$$
$$=\int\frac{4u^2+2u+2}{u^2+\frac{1}{2}}du$$
$$=\int\frac{4(u^2+\frac{1}{2})}{u^2+\frac{1}{2}}+\int\frac{2u}{u^2+\frac{1}{2}}du$$
$$=4u + \ln|u^2+\frac{1}{2}|+c$$

Substituting back

$$\int\frac{4x^{2}-2x+2}{4x^{2}-4x+3}dx=4(x-\frac{1}{2})+\ln|(x-\frac{1}{2})^2+\frac{1}{2})|+c$$
$$=4(x-\frac{1}{2})+\ln|4x^{2}-4x+3|+c$$

Would someone be so kind as to tell me if this is correct? For this question, I have 4 possible answers (and one that states "None of the above") and my answer doesn't match any of the other 3, so I'm wondering.

Thanks!
phyz

2. Jul 5, 2008

### rocomath

I won't say if it's good or not since it seems like you're taking a quiz/exam or something. Differentiate to check your original Integral.

3. Jul 5, 2008

### HallsofIvy

It seems to me you could simplify the problem by long division before completing the square.

4. Jul 5, 2008

### Astronuc

Staff Emeritus
$$\int\frac{4x^{2}-2x+2}{(x-\frac{1}{2})^2+\frac{1}{2}}dx$$

What happened to the factor of 4 in the denominator?

5. Jul 6, 2008

### phyzmatix

Close...I'm busy with an assignment, but the whole thing only contributes about 5% to my year mark and is more for tutorial purposes than anything else since, where I'm studying, we don't have lectures at all.

Thanks HallsofIvy, I'm going to give that a try today.

Mmmm...Oh, wait, let me show you (perhaps I did something wrong here as well)

$$4x^2-4x+3 = 0$$
$$4x^2-4x=-3$$
$$x^2-x=-\frac{3}{4}$$
$$x^2-x+(\frac{1}{2})^2=-\frac{3}{4}+(\frac{1}{2})^2$$
$$(x-\frac{1}{2})^2+\frac{1}{2}=0$$

?

Last edited: Jul 6, 2008
6. Jul 6, 2008

### dirk_mec1

$$\int \frac{4x^{2}-2x+2}{4x^{2}-4x+3} \mbox{d}x = \int \frac{1}{4x^{2}-4x+3}\ +\ \frac{2x-1}{4x^{2}-4x+3}\ \mbox{d}x=....$$

7. Jul 7, 2008

### phyzmatix

[SOLVED]

PS. dirk_mec1, I think it should be

$$\int 1 dx + \int\frac{2x-1}{4x^2-4x+3}dx$$

8. Jul 7, 2008

### HallsofIvy

???
$$\frac{1}{4x^{2}-4x+3}\ +\ \frac{2x-1}{4x^{2}-4x+3}= \frac{2x}{4x^2- 4x+ 3}$$

Did you mean
$$\int \frac{4x^{2}-2x+2}{4x^{2}-4x+3} \mbox{d}x = \int 1\ +\ \frac{2x-1}{4x^{2}-4x+3}\ \mbox{d}x$$

9. Jul 7, 2008

### dirk_mec1

Yes sorry :shy: