# How do you use partial fractions to find the integral #int (x^2-x+2)/(x^3-x^2+x-1)dx#?

Then, we can write a systems of equations.

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To use partial fractions to find the integral (\int \frac{x^2-x+2}{x^3-x^2+x-1}dx), first factor the denominator as (x^3-x^2+x-1 = (x-1)(x^2+1)).

The denominator has a linear factor (x-1) and an irreducible quadratic factor (x^2+1). Therefore, we can write the integrand as:

(\frac{x^2-x+2}{x^3-x^2+x-1} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}).

To find the values of (A), (B), and (C), multiply both sides by the denominator (x^3-x^2+x-1) to get:

(x^2 - x + 2 = A(x^2 + 1) + (Bx + C)(x - 1)).

Now, substitute convenient values of (x) to solve for (A), (B), and (C). For example, setting (x = 1) gives:

(2 = A(1^2 + 1) \Rightarrow 2 = 2A \Rightarrow A = 1).

Setting (x = 0) gives:

(2 = A(0^2 + 1) + C(-1) \Rightarrow 2 = C(-1) \Rightarrow C = -2).

Finally, setting (x = i) (where (i) is the imaginary unit) gives:

(-i = A(i^2 + 1) + (Bi - 2)(i - 1)).

Simplify and solve for (B):

(-i = A(-1 + 1) + (Bi - 2)(i - 1)) (-i = (Bi - 2)(i - 1)) (-i = Bi^2 - Bi - 2i + 2) (-i = -B - 2i + 2).

Equating real and imaginary parts gives:

(-B + 2 = 0) and (-2 = -2).

Thus, (B = 2).

Now, the integral becomes:

(\int \frac{x^2-x+2}{x^3-x^2+x-1}dx = \int \frac{1}{x-1}dx + \int \frac{2x-2}{x^2+1}dx).

Integrating these terms separately gives:

(\int \frac{1}{x-1}dx = \ln|x-1| + C_1),

(\int \frac{2x-2}{x^2+1}dx = \int \frac{2x}{x^2+1}dx - \int \frac{2}{x^2+1}dx = \ln|x^2+1| - 2\arctan(x) + C_2).

Therefore, the final result is:

(\int \frac{x^2-x+2}{x^3-x^2+x-1}dx = \ln|x-1| + \ln|x^2+1| - 2\arctan(x) + C), where (C = C_1 + C_2).

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