Evaluate the integral: I = integral (x²)/(x²-3x+2) dx

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

Explanation of Method

Since the numerator and denominator have equal degree, polynomial long division peels off a constant term first; partial fractions handle what remains.

Divide $x^2$ by $x^2-3x+2$.
Factor the denominator and decompose the remainder into partial fractions.
Integrate term by term.

Step 1 · Polynomial long division

$$ \frac{x^2}{x^2-3x+2} = 1 + \frac{3x-2}{x^2-3x+2}. $$

Step 2 · Factor and decompose the remainder

$$ x^2-3x+2=(x-1)(x-2), \qquad \frac{3x-2}{(x-1)(x-2)} = \frac{A}{x-1}+\frac{B}{x-2}. $$

Multiplying through: $3x-2=A(x-2)+B(x-1)$. Setting $x=1$: $1=-A$, so $A=-1$. Setting $x=2$: $4=B$.

Step 3 · Integrate

$$ I = \int 1\,dx + \int \left(\frac{-1}{x-1}+\frac{4}{x-2}\right) dx = x - \ln|x-1| + 4\ln|x-2| + C. $$

Step 4 · Combine logarithms

$$ I = x + \ln\left(\frac{(x-2)^4}{|x-1|}\right) + C. $$

Summary Table

Coefficient	Value
$A$	$-1$
$B$	$4$

Additional Notes

The domain excludes $x=1$ and $x=2$, the roots of the denominator.
Since the numerator's degree equals the denominator's, the division quotient is just a constant ($1$), not a full polynomial — a simpler special case of the technique used in Problems 231–236.
The result can be checked by differentiating $x-\ln|x-1|+4\ln|x-2|$: this gives $1-\tfrac{1}{x-1}+\tfrac{4}{x-2}$; combining the fractions over $(x-1)(x-2)$ gives $1+\tfrac{3x-2}{x^2-3x+2}=\tfrac{(x^2-3x+2)+(3x-2)}{x^2-3x+2}=\tfrac{x^2}{x^2-3x+2}$, matching the integrand.

Result

The solution to the integral is:

$$ I = x + \ln\left(\frac{(x-2)^4}{|x-1|}\right) + C. $$