Mastering Definite Integrals for Solving Tricky Problems

sinClair
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Homework Statement


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

Homework Equations

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

sinClair said:

Homework Statement


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


Homework Equations




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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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