-*-dx)%2F(sqrt(x-%5E-2---3x-%2B-2)).jpeg "((3x ^ 3 - X ^ 2 + 2x - 4) * Dx)/(sqrt(x ^ 2 - 3x + 2))")
Evaluating the Integral of a Rational Function
In this article, we will evaluate the integral of a rational function:
$\int \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} dx$
Step 1: Simplify the Denominator
The first step is to simplify the denominator by factoring the quadratic expression inside the square root:
$x^2 - 3x + 2 = (x - 1)(x - 2)$
So, the integral becomes:
$\int \frac{3x^3 - x^2 + 2x - 4}{\sqrt{(x - 1)(x - 2)}} dx$
Step 2: Substitute u = x - 1
To simplify the integral, we can substitute $u = x - 1$. Then, $du = dx$ and $x = u + 1$.
Substituting these values into the integral, we get:
$\int \frac{3(u + 1)^3 - (u + 1)^2 + 2(u + 1) - 4}{\sqrt{u(u - 1)}} du$
Step 3: Simplify the Numerator
Expanding the numerator, we get:
$3u^3 + 6u^2 + 3u + 1 - u^2 - 2u - 1 + 2u + 2 - 4$
Simplifying the numerator, we get:
$3u^3 + 5u^2 + u - 2$
Step 4: Evaluate the Integral
Now, we can evaluate the integral:
$\int \frac{3u^3 + 5u^2 + u - 2}{\sqrt{u(u - 1)}} du$
This is a standard integral, and we can evaluate it using integration by partial fractions.
Solution
After evaluating the integral, we get:
$\frac{2}{3}u^{3/2} + \frac{5}{2}u^{1/2} + u^{-1/2} + C$
Substituting back $u = x - 1$, we get:
$\frac{2}{3}(x - 1)^{3/2} + \frac{5}{2}(x - 1)^{1/2} + (x - 1)^{-1/2} + C$
This is the final solution to the integral.
Conclusion
In this article, we have evaluated the integral of a rational function using substitution and integration by partial fractions. The final solution is a combination of power functions and is defined for $x > 2$.
