$\displaystyle
\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
$\displaystyle
\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
I think it's a pretty safe bet to say he did the above by practicing high school algebra.
Now, $\displaystyle \frac{6x - 6}{x^2 - 2x + 2} = 3\frac{2x - 2}{x^2 - 2x + 2}$ and this is a logarithmic integral since the numerator is the denominator's derivative
Likewise the second summand is a 3Arctan(x-1)
Tonio