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!