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. $\int \frac{1}{{(x-\frac{1}{2})}^2+\frac{1}{4}}dx$
• B. $\int \frac{1}{{(x-\frac{1}{2})}^2+\frac{3}{4}}dx$
• C. $\int \frac{1}{{(x-\frac{1}{2})}^2+\frac{1}{2}}dx$

Answer 1

The correct option is B $\int \frac{1}{{(x-\frac{1}{2})}^2+\frac{3}{4}}dx$

The denominator is $x^2-x+1$
This can be written as $x^2-2\times \frac{1}{2}\times x+1$
This is of the form $a^2+2ab+c$. We need to add and subtract $b^2$
Thus, $x^2-2\times \frac{1}{2}\times x+1=x^2-2\times \frac{1}{2}\times x+{\left(\frac{1}{2}\right)}^2+1-{\left(\frac{1}{2}\right)}^2$
$=x^2-2\times \frac{1}{2}\times x+\frac{1}{4}+1-\frac{1}{4}$
$={(x-\frac{1}{2})}^2+\frac{3}{4}$

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