(3x^3-x^2+2x-4)dx
--------------------------
sqrt x^2-3x+2
the answer is supposed to be between 60000 and 70000
Is this supposed to be some integral?
If you want a constant answer you need a lower and upper value.
I gave the algebraic expression to Wolfram, don't know if this helps.
(I certainly would not know how to integrate this right now)
http://www.wolframalpha.com/input/?i=integral+(3x%5E3-x%5E2%2B2x-4)%2F%E2%88%9A(x%5E2-3x%2B2)dx
To find the integral of the function
(3x^3 - x^2 + 2x - 4) / sqrt(x^2 - 3x + 2),
we can use a combination of algebraic manipulation and integration techniques.
Step 1: Factoring the denominator
To make the integration process easier, we should start by factoring the denominator as much as possible.
The denominator, sqrt(x^2 - 3x + 2), can be written as sqrt((x - 2)(x - 1)).
Step 2: Decomposing into partial fractions
Next, we decompose the rational function into partial fractions.
The rational function can be rewritten as
(3x^3 - x^2 + 2x - 4) / sqrt((x - 2)(x - 1)).
We need to determine the unknown coefficients A, B, C, and D.
To do this, we multiply both sides of the equation by the denominator, yielding:
(3x^3 - x^2 + 2x - 4) = A * sqrt(x - 2) + B * sqrt(x - 1) + C * (x - 2) + D * (x - 1).
Now, we can solve this equation for A, B, C, and D.
Step 3: Integrating the partial fractions
Once we have determined A, B, C, and D, we can integrate each term separately.
Integrating A * sqrt(x - 2) gives 2/3 * (x - 2)^1.5.
Integrating B * sqrt(x - 1) gives 2/3 * (x - 1)^1.5.
Integrating C * (x - 2) gives 0.5 * C * (x - 2)^2.
Integrating D * (x - 1) gives 0.5 * D * (x - 1)^2.
Step 4: Combining the integrals
Finally, we add up the integrals of the partial fractions to get the integral of the original rational function.
The integral of (3x^3 - x^2 + 2x - 4) / sqrt(x^2 - 3x + 2) is:
(2/3 * (x - 2)^1.5) + (2/3 * (x - 1)^1.5) + (0.5 * C * (x - 2)^2) + (0.5 * D * (x - 1)^2) + C1,
where C1 is the constant of integration.