How do you integrate #(x) / (x^(3)-x^(2)-2x +2)# using partial fractions?

Question

Christopher Allers · Accepted Answer

#=\ln|\frac{\sqrt{x^2-2}}{x-1}|+1/{\sqrt2}\ln|\frac{x-\sqrt2}{x+\sqrt2}|+C#

Making partial fractions as follows

#\frac{x}{x^3-x^2-2x+2}#

#=\frac{x}{(x-1)(x^2-2)}#

#=A/(x-1)+{Bx+C}/{x^2-2}#

Comparing corresponding coefficients on both sides we get

#A=-1, B=1, C=2#

Now,

#\frac{x}{x^3-x^2-2x+2}=-1/{x-1}+{x+2}/{x^2-2}#

Integrating above equation on both the sides w.r.t. #x#, we get

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

#=-\int 1/{x-1}\ dx+\int x/{x^2-2}\ dx+2\int 1/{x^2-2}\ dx#

#=-ln|x-1|+1/2\int {d(x^2-2)}/{x^2-2}\ dx+2\int 1/{x^2-(\sqrt2)^2}\ dx#

#=-ln|x-1|+1/2\ln|x^2-2|+2\cdot 1/{2\sqrt2}\ln|\frac{x-\sqrt2}{x+\sqrt2}|+C#

#=\ln|\frac{\sqrt{x^2-2}}{x-1}|+1/{\sqrt2}\ln|\frac{x-\sqrt2}{x+\sqrt2}|+C#

Axel Alvarez · Answer

undefined To integrate $\frac{x}{x^3-x^2-2x+2}$ using partial fractions, follow these steps:

1. Factor the denominator: $x^3-x^2-2x+2 = (x-1)(x^2-2)$.
2. Express the fraction as the sum of two partial fractions: $\frac{x}{(x-1)(x^2-2)} = \frac{A}{x-1} + \frac{Bx + C}{x^2-2}$.
3. Clear the denominators by multiplying both sides by $(x-1)(x^2-2)$.
4. Simplify and solve for $A$, $B$, and $C$ by comparing coefficients.
5. Once you have the values of $A$, $B$, and $C$, rewrite the fraction in terms of the partial fractions.
6. Integrate each partial fraction separately.
7. Combine the results to obtain the final integrated expression.