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

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Answer 1 · 2016-12-17T20:53:45

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

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!

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