Integral of rational function, could I use efficient way?

Question

I have the integral $$\int \frac{x^3+1}{x^3-5x^2+6x}dx$$ I chose to solve it in the following way : first split it for sum of two fractions and decrease the degree of the polynomial of the numerator. $$\int \frac{x^3+1}{x^3-5x^2+6x}\,dx= \int \frac{x^3}{x^3-5x^2+6x}\,dx +\int \frac{+1}{x^3-5x^2+6x}\,dx $$ \begin{align} \int \frac{x^3}{x^3-5x^2+6x}&\,dx=\int \frac{x^3+5x^2 -5x^2+6x -6x}{x^3-5x^2+6x}\,dx \\ = x + {}&\int \frac{5x^2-6x}{x^3-5x^2+6x}\,dx = x + \int \frac{5x-6}{x^2-5x+6}\,dx\\ = x +{} &\int \frac{-4}{x-2}\,dx +\int \frac{9}{x-3}\,dx = x -4\ln|x-2| + 9\ln|x-3| \end{align}

the second integral I solved in similar way and after sum them both I got the answer $$x+ \frac{1}{6}\ln|x| - \frac{9}{2}\ln|x-2| + \frac{28}{3}\ln|x-3| + c $$ wich is correct answer. But my question is if I could solve it in much shortest and efficient way? could I use long polynomial long dividing?

Answer 1

You don't need to split $x^3+1$ at the beginning. Performing long division you get that $$ \frac{x^3+1}{x^3-5x^2+6x}=1+\frac{5x^2-6x+1}{x^3-5x^2+6x} $$ Then write the denominator as $x(x-2)(x-3)$ and you can rewrite the fraction as $$ \frac{A}{x}+\frac{B}{x-2}+\frac{C}{x-3}=\frac{5x^2-6x+1}{x^3-5x^2+6x} $$ and you get a system in $A,B,C$ by comparing the coefficients. $$ A+B+C=5 \\ -5A-3B-2C=-6\\ 6A=1 $$

So, after you found $A,B,C$, the solution is given by $$ x+A \ln|x|+B \ln |x-2|+C\ln|x-3|+c $$

Answer 2

Much shorter, I don't know. But the standard way consists in obtaining first the decomposition into partial fractions, which is valid for proper rational fractions i.e. fractions $\dfrac{P(x)}{Q(x)}$ s.t. $\deg P<\deg Q$.

• When it is not the case, we first perform the Euclidean divieion of $P(x)$ by $Q(x)$, which is trivial here: $$\frac{x^3+1}{x^3-5x^2+6x}=\frac{(x^3-5x^2+6x)+5x^2-6x+1}{x^3-5x^2+6x}=x+\frac{5x^2-6x+1}{x^3-5x^2+6x}$$
• Next we write the denominator as a product of irreducible factors: $$x^3-5x^2+6x=x(x-2)(x-3).$$
• So we have a decomposition of the proper rational fraction $$\frac{5x^2-6x+1}{x^3-5x^2+6x}=\frac Ax+\frac B{x-2}+\frac C{x-3}$$ As the roots are simple roots, there is a formula for the coefficients: denoting in a generic way the numerator and denominator as $P(x)$ and $Q(x)$, we have $$A=\frac{P(0)}{Q'(0)}=\frac16,\qquad B=\frac{P(2)}{Q'(2)}=-\frac 92,\qquad C=\frac{P(3)}{Q'(3)}=\frac{28}3.$$