Question

The form of the partial fraction decomposition of a rational function is given below.

5x²-2x+22/(x-1)(x²+4) = A/x-1 + Bx+c/x²+4
A = __
B = __
C = __
Now evaluate the indefinite integral. ∫5x²-2x+22/(x-1)(x²+4) dx =

Answer 1

The values of A, B, and C in the given partial fraction decomposition are: A = -3, B = 5, and C = -10. The indefinite integral of the rational function is: ∫(5x² - 2x + 22)/((x - 1)(x² + 4)) dx = -3ln|x - 1| + 5ln|x² + 4| - 10arctan(x/2) + C

Step-by-step explanation:

To find the values of A, B, and C in the given partial fraction decomposition, let's start by setting up the equation:

5x² - 2x + 22 = A(x² + 4) + (Bx + C)(x - 1)

Expanding and collecting like terms, we get:

5x² - 2x + 22 = Ax² + 4A + Bx² - Bx + Cx - C

Next, let's match the coefficients of the same powers of x on both sides of the equation:

For x²: 5A + B = 5

For x: -2 - B + C = 0

For constant term: 4A - C + 22 = 0

Solving these equations simultaneously, we find that A = -3, B = 5, and C = -10.

Now, to evaluate the indefinite integral, we can use the partial fraction decomposition:

∫(5x² - 2x + 22)/((x - 1)(x² + 4)) dx = ∫(-3/(x - 1) + (5x - 10)/(x² + 4)) dx

Integrating each term separately, we get:

-3ln|x - 1| + 5ln|x² + 4| - 10arctan(x/2) + C