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

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Answer 1 · 2018-07-19T14:52:13

#=\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#