Question

integral 0,1 (3x^2 -x^2 +2x -4)/(√(x^2 - 3x +2)) dx

Answer 1

I think the question is $\bold{\int\limits^1_0{\frac{(3x^3- x^2 +2x-4)}{(x^2-3x+2)}}\,dx}$
First of all factorize numerator and denominator.
e.g., numerator : 3x³ - x² + 2x - 4
= 3x³ - 3x² + 2x² - 2x + 4x - 4
= (x - 1)(3x² + 2x + 4)

denominator : x² - 3x + 2
= x² - 2x - x + 2
= (x - 2)(x - 1)

So, integration converts into $\bold{\int\limits^1_0{\frac{(x-1)(3x^2+2x+4)}{(x-1)(x -2)}}\,dx}$
Or, $\bold{\int\limits^1_0{\frac{(3x^2+2x+4)}{(x -2)}}\,dx}$
Now, see attachment.....

Answer 2

Answer:

step by step. explanation