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

Dec 17, 2016

$\ln | x - 1 | - \arctan x + C$

#### Explanation:

The denominator can be factored as ${x}^{2} \left(x - 1\right) + 1 \left(x - 1\right) = \left({x}^{2} + 1\right) \left(x - 1\right)$.

$\frac{A x + B}{{x}^{2} + 1} + \frac{C}{x - 1} = \frac{{x}^{2} - x + 2}{\left({x}^{2} + 1\right) \left(x - 1\right)}$

$\left(A x + B\right) \left(x - 1\right) + C \left({x}^{2} + 1\right) = {x}^{2} - x + 2$

$A {x}^{2} + B x - A x - B + C {x}^{2} + C = {x}^{2} - x + 2$

$\left(A + C\right) {x}^{2} + \left(B - A\right) x + \left(C - B\right) = {x}^{2} - x + 2$

Then, we can write a systems of equations.

$\left\{\begin{matrix}A + C = 1 \\ B - A = - 1 \\ C - B = 2\end{matrix}\right.$

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

Therefore, the partial fraction decomposition is $- \frac{1}{{x}^{2} + 1} + \frac{1}{x - 1}$.

The integral becomes $\int \left(\frac{1}{x - 1} - \frac{1}{{x}^{2} + 1}\right) \mathrm{dx}$.

Note that $\frac{d}{\mathrm{dx}} \left(\arctan x\right) = \frac{1}{{x}^{2} + 1} \mathrm{dx}$ and that $\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x} \mathrm{dx}$.

Therefore, the integral is $\ln | x - 1 | - \arctan x + C$.

Hopefully this helps!