Evaluate the integral: I = integral (x³+x²)/(x²-3x+2) dx

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

Step 1 · Polynomial long division

$$ \frac{x^3+x^2}{x^2-3x+2} = x+4+\frac{10x-8}{x^2-3x+2}. $$

Step 2 · Integrate the polynomial part

$$ \int (x+4)\,dx = \frac{x^2}{2}+4x. $$

Step 3 · Factor and decompose the remainder

$$ x^2-3x+2=(x-1)(x-2), \qquad \frac{10x-8}{(x-1)(x-2)} = \frac{A}{x-1}+\frac{B}{x-2}. $$

Multiplying through: $10x-8=A(x-2)+B(x-1)$. Setting $x=1$: $2=-A$, so $A=-2$. Setting $x=2$: $12=B$.

Step 4 · Integrate and combine

$$ \int \frac{10x-8}{x^2-3x+2}\,dx = -2\ln|x-1|+12\ln|x-2|. $$

$$ I = \frac{x^2}{2}+4x - 2\ln|x-1| + 12\ln|x-2| + C. $$

Summary Table

Additional Notes

• The domain excludes $x=1$ and $x=2$, the roots of the denominator.
• This closely parallels Problem 233, sharing the same denominator $x^2-3x+2$; only the numerator (and hence the division quotient and partial-fraction coefficients) differs.
• The result can be checked by differentiating $\tfrac{x^2}2+4x-2\ln|x-1|+12\ln|x-2|$: this gives $x+4-\tfrac{2}{x-1}+\tfrac{12}{x-2}=x+4+\tfrac{10x-8}{x^2-3x+2}=\tfrac{(x+4)(x^2-3x+2)+10x-8}{x^2-3x+2}$; since $(x+4)(x^2-3x+2)=x^3+x^2-10x+8$, adding $10x-8$ gives $x^3+x^2$, matching the integrand.