# Confusing Definite Integral

## Homework Statement

Integrate $$\int_{0}^{1}\sqrt{\frac{4x^2-4x+1}{x^2-x+3}}dx$$

## The Attempt at a Solution

U sub: let $$u=x^2-x+3$$ Then $$du=2x-1$$ and then have to evaluate $$\int_{3}^{3}\sqrt{\frac{du^2}{u}}dx$$ But how with these limits of integration should this be 0? Not sure how to evaluate this....

Complete the square in both numerator and denominator, then use a u-sub for the radican.

HallsofIvy
Homework Helper
Complete the square in both numerator and denominator, then use a u-sub for the radican.
I see no reason to do that.

## Homework Statement

Integrate $$\int_{0}^{1}\sqrt{\frac{4x^2-4x+1}{x^2-x+3}}dx$$

## The Attempt at a Solution

U sub: let $$u=x^2-x+3$$ Then $$du=2x-1$$ and then have to evaluate $$\int_{3}^{3}\sqrt{\frac{du^2}{u}}dx$$ But how with these limits of integration should this be 0? Not sure how to evaluate this....
$\sqrt{x^2}= |x|$ and 2x- 1 changes sign at x= 1/2. This integral is the same as
$$\int_0^1 \frac{|2x-1|}{\sqrt{x^2- x+ 3}}dx= -\int_0^{1/2}\frac{2x-1}{\sqrt{x^2-x+ 3}}dx+ \int_{1/2}^1 \frac{2x-1}{\sqrt{x^2- x+ 3}}dx$$

Thank you so much. I was getting confused with changing the limits of integration back and forth but got it now, thanks.