Evaluate the integral: I = integral (5x-8)/(x³-4x²+4x) dx

Evaluate the integral: $$ I = \int \frac{5x-8}{x^3-4x^2+4x}\,dx. $$

Step 1 · Factor and set up partial fractions

$$ x^3-4x^2+4x = x(x-2)^2, \qquad \frac{5x-8}{x(x-2)^2} = \frac Ax + \frac{B}{x-2} + \frac{C}{(x-2)^2}. $$

Multiplying through: $5x-8=A(x-2)^2+Bx(x-2)+Cx$.

Step 2 · Solve for the coefficients

Setting $x=0$: $-8=4A$, so $A=-2$. Setting $x=2$: $2=2C$, so $C=1$. Setting $x=1$: $-3=A-B+C=-2-B+1$, giving $B=2$.

Step 3 · Integrate each term

$$ I = \int \left(\frac{-2}{x} + \frac{2}{x-2} + \frac{1}{(x-2)^2}\right) dx = -2\ln|x| + 2\ln|x-2| - \frac{1}{x-2} + C. $$

Summary Table

Additional Notes

• The domain excludes $x=0$ and $x=2$, the roots of the denominator.
• The two logarithms combine as $2\ln|x-2|-2\ln|x|=2\ln\left|\tfrac{x-2}{x}\right|$.
• The result can be checked by differentiating $-2\ln|x|+2\ln|x-2|-\tfrac{1}{x-2}$: this gives $-\tfrac2x+\tfrac2{x-2}+\tfrac1{(x-2)^2}$, and combining over $x(x-2)^2$ gives numerator $-2(x-2)^2+2x(x-2)+x$, which expands to $5x-8$, matching the integrand.