Question

If we convert the denominator of the integral into a perfect square, $\int \frac{1}{x^2-x+1}$dx then the correct integral will be

• A. dx
• B. dx
• C. dx
• D.

Answer 1

The correct option is B dx

We know that $a{x}^2+bx+c$ can also be written as a $\left[\right(x+b/2a{)}^2+c/a-b^2/4a^2]$ by completing squares, which we learnt in quadratic equations. We'll apply the same formula here. In the question given here, a = 1, b = -1, c = 1
So, $x^2-x+1=1\left[\right(x-1/2{)}^2+1-1/4]$
or $x^2-x+1=\left[\right(x-1/2{)}^2+3/4]$
The given integral $\int \frac{1}{x^2-x+1}$dx can be written as $textstyle\int \frac{1}{(x-1/2{)}^2+3/4}$dx