Integral of a fraction consisting of two quadratic equations

phyzmatix

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

[tex]\int\frac{4x^{2}-2x+2}{4x^{2}-4x+3}dx[/tex]


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


3. The attempt at a solution
Completing the square for

[tex]4x^{2}-4x+3[/tex]

gives

[tex]\int\frac{4x^{2}-2x+2}{(x-\frac{1}{2})^2+\frac{1}{2}}dx[/tex]

let

[tex]u=x-\frac{1}{2}, du = dx, x=u+\frac{1}{2}[/tex]

then

[tex]\int\frac{4(u+\frac{1}{2})^{2}-2(u+\frac{1}{2})+2}{u^2+\frac{1}{2}}du[/tex]
[tex]=\int\frac{4u^2+4u+1-2u-1+2}{u^2+\frac{1}{2}}du[/tex]
[tex]=\int\frac{4u^2+2u+2}{u^2+\frac{1}{2}}du[/tex]
[tex]=\int\frac{4(u^2+\frac{1}{2})}{u^2+\frac{1}{2}}+\int\frac{2u}{u^2+\frac {1}{2}}du[/tex]
[tex]=4u + \ln|u^2+\frac{1}{2}|+c[/tex]

Substituting back

[tex]\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[/tex]
[tex]=4(x-\frac{1}{2})+\ln|4x^{2}-4x+3|+c[/tex]

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

[tex]\int\frac{4x^{2}-2x+2}{(x-\frac{1}{2})^2+\frac{1}{2}}dx[/tex]

What happened to the factor of 4 in the denominator?

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)

[tex]4x^2-4x+3 = 0[/tex]
[tex]4x^2-4x=-3[/tex]
[tex]x^2-x=-\frac{3}{4}[/tex]
[tex]x^2-x+(\frac{1}{2})^2=-\frac{3}{4}+(\frac{1}{2})^2[/tex]
[tex](x-\frac{1}{2})^2+\frac{1}{2}=0[/tex]

?

dirk_mec1

[tex]
\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=....
[/tex]

phyzmatix

[SOLVED]



PS. dirk_mec1, I think it should be

[tex]\int 1 dx + \int\frac{2x-1}{4x^2-4x+3}dx[/tex]

HallsofIvy

???
[tex]\frac{1}{4x^{2}-4x+3}\ +\ \frac{2x-1}{4x^{2}-4x+3}= \frac{2x}{4x^2- 4x+ 3}[/tex]

Did you mean
[tex]
\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
[/tex]

dirk_mec1

Yes sorry