Integral of a fraction consisting of two quadratic equations

[phyzmatix]

Homework Statement

Determine

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

Homework Equations

I believe it's necessary to complete the square.

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

 

[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.

 

[HallsofIvy]

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

 

[Astronuc]

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

What happened to the factor of 4 in the denominator?

 

[phyzmatix]

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.

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.

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

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

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

What happened to the factor of 4 in the denominator?

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

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

?

 

[dirk_mec1]

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

 

[phyzmatix]

[SOLVED]



PS. dirk_mec1, I think it should be

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

 

[HallsofIvy]

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

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

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

 

[dirk_mec1]

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

Yes sorry :shy: