Question

∫ (3׳-X²+2×-4)dx √x²-3x+2

Answer 1

=3*x⁴/4-x³/3+2x²/2-4x=
3x⁴/4-x³/3+x²-4x

Answer 2

To integrate the given expression, let's first factor the denominator under the square root (√):

√(x² - 3x + 2) = √((x - 2)(x - 1))

Now, we can express the given integral as a sum of three integrals:

∫ [(3x³ - x² + 2x - 4) / √(x² - 3x + 2)] dx = ∫ (3x³ - x² + 2x - 4) / √((x - 2)(x - 1)) dx

Now, we'll use partial fraction decomposition to break down the integrand into simpler fractions:

(3x³ - x² + 2x - 4) / √((x - 2)(x - 1)) = A / √(x - 2) + B / √(x - 1)

Now, we need to find the values of A and B. To do this, let's find a common denominator, which is the square root of the product of the denominators:

Common denominator = √(x - 2) * √(x - 1) = √((x - 2)(x - 1))

Now, we can write the original expression with the common denominator:

(3x³ - x² + 2x - 4) / √((x - 2)(x - 1)) = (A * √(x - 1) + B * √(x - 2)) / √((x - 2)(x - 1))

To find the values of A and B, we equate the numerators:

3x³ - x² + 2x - 4 = A * √(x - 1) + B * √(x - 2)

Now, let's find A and B by comparing coefficients:

1. Coefficient of √(x - 1): On the left side, there is no term with √(x - 1), so the coefficient of √(x - 1) on the right side is 0. A * √(x - 1) = 0 A = 0

2. Coefficient of √(x - 2): On the left side, there is no term with √(x - 2), so the coefficient of √(x - 2) on the right side is also 0. B * √(x - 2) = 0 B = 0

Now, the integral becomes:

∫ [(3x³ - x² + 2x - 4) / √(x² - 3x + 2)] dx = ∫ (A / √(x - 2) + B / √(x - 1)) dx = ∫ (0 / √(x - 2) + 0 / √(x - 1)) dx = 0

Therefore, the integral of the given expression is 0.