Evaluate the integral: I = integral (x³+1)/(x²-3x+2) dx

Evaluate the integral: $$ I = \int \frac{x^3+1}{x^2-3x+2}\,dx. $$

Step 1 · Polynomial long division

$$ \frac{x^3+1}{x^2-3x+2} = x+3+\frac{7x-5}{x^2-3x+2}. $$

Step 2 · Integrate the polynomial part

$$ \int (x+3)\,dx = \frac{x^2}{2}+3x. $$

Step 3 · Factor and decompose the remainder

$$ x^2-3x+2=(x-1)(x-2), \qquad \frac{7x-5}{(x-1)(x-2)} = \frac{A}{x-1}+\frac{B}{x-2}. $$

Multiplying through: $7x-5=A(x-2)+B(x-1)$. Setting $x=1$: $2=-A$, so $A=-2$. Setting $x=2$: $9=B$.

Step 4 · Integrate and combine

$$ \int \frac{7x-5}{x^2-3x+2}\,dx = \int \left(\frac{-2}{x-1}+\frac{9}{x-2}\right) dx = -2\ln|x-1|+9\ln|x-2|. $$

$$ I = \frac{x^2}{2}+3x - 2\ln|x-1| + 9\ln|x-2| + C. $$

Summary Table

Additional Notes

• The domain excludes $x=1$ and $x=2$, the roots of the denominator.
• The cover-up method (plugging each root directly into the reduced fraction) is faster and less error-prone here than splitting $7x-5$ into a multiple of the denominator's derivative plus a constant, which requires an extra algebraic step.
• The result can be checked by differentiating $\tfrac{x^2}2+3x-2\ln|x-1|+9\ln|x-2|$: this gives $x+3-\tfrac{2}{x-1}+\tfrac{9}{x-2}=x+3+\tfrac{7x-5}{x^2-3x+2}=\tfrac{(x+3)(x^2-3x+2)+7x-5}{x^2-3x+2}=\tfrac{x^3+1}{x^2-3x+2}$, matching the integrand (since $(x+3)(x^2-3x+2)=x^3-7x+6$, and adding $7x-5$ gives $x^3+1$).