1. Integrating partial fractions

$

\int (6x - 3)/(x^2 - 2x + 2) dx
$

I was wondering if I could get some help on this question. Would you use completing the square for the denominator, since it can't be readily factorised? If so, how exactly?

Thanks

2. Originally Posted by Danog
$

\int (6x - 3)/(x^2 - 2x + 2) dx
$

I was wondering if I could get some help on this question. Would you use completing the square for the denominator, since it can't be readily factorised? If so, how exactly?

Thanks
[/font]
You need to write the integrand as $\frac{6x - 6 + 3}{x^2 - 2x + 2} = \frac{6x - 6}{x^2 - 2x + 2} + \frac{3}{x^2 - 2x + 2} = \frac{6x - 6}{x^2 - 2x + 2} + \frac{3}{(x - 1)^2 + 1}$ and it should be apparent how to integrate each of these.

3. Originally Posted by mr fantastic
You need to write the integrand as $\frac{6x - 6 + 3}{x^2 - 2x + 2} = \frac{6x - 6}{x^2 - 2x + 2} + \frac{3}{x^2 - 2x + 2} = \frac{6x - 6}{x^2 - 2x + 2} + \frac{3}{(x - 1)^2 + 1}$ and it should be apparent how to integrate each of these.
I'm still a bit confused. How do you know to do that? What is the next step? I've been I'll for an extended period so I'm not entirely sure of what to do.

Thanks again

4. $\int \frac{6x - 6}{x^2 - 2x + 2} + \frac{3}{(x - 1)^2 + 1} \, dx$

$\int 3 \frac{2x - 2}{x^2 - 2x + 2} + \frac{3}{(x - 1)^2 + 1} \, dx$

$3\ln(x^2-2x+2) + 3\arctan(x-1) + C$

5. Originally Posted by Danog
I'm still a bit confused. How do you know to do that? What is the next step? I've been I'll for an extended period so I'm not entirely sure of what to do.

Thanks again
I think it's a pretty safe bet to say he did the above by practicing high school algebra.

Now, $\frac{6x - 6}{x^2 - 2x + 2} = 3\frac{2x - 2}{x^2 - 2x + 2}$ and this is a logarithmic integral since the numerator is the denominator's derivative

Likewise the second summand is a 3Arctan(x-1)

Tonio