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

Question

Ezekiel Alexander · Accepted Answer

#ln|x - 1| - arctanx + C#

The denominator can be factored as #x^2(x - 1) + 1(x- 1) = (x^2 + 1)(x- 1)#.

#(Ax + B)/(x^2 + 1) + C/(x- 1) = (x^2 - x + 2)/((x^2 + 1)(x - 1))#

#(Ax + B)(x- 1) + C(x^2 + 1) = x^2 - x + 2#

#Ax^2 + Bx - Ax - B + Cx^2 + C = x^2 - x + 2#

#(A + C)x^2 + (B - A)x + (C - B) = x^2 - x + 2#

Then, we can write a systems of equations.

#{(A + C = 1), (B - A = -1), (C - B = 2):}#

Solve to get #A = 0, B = -1, C= 1#.

Therefore, the partial fraction decomposition is #-1/(x^2 + 1) + 1/(x - 1)#.

The integral becomes #int(1/(x - 1) - 1/(x^2 + 1))dx#.

Note that #d/dx(arctanx) = 1/(x^2 + 1)dx# and that #d/dx(lnx) = 1/xdx#.

Therefore, the integral is #ln|x - 1| - arctanx + C#.

Hopefully this helps!

Christopher Agresta · Answer

undefined 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 $C = C_1 + C_2$.