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. $\int \frac{1}{(x-1/2{)}^2+1}dx$
• B. $\int \frac{1}{(x-1/2{)}^2+3/4}dx$
• C. $\int \frac{1}{(x-1{)}^2+1/2}dx$
• D. None of these

Answer 1

The correct option is B

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

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