Integrals of simple fractions

• The degree of the numerator is less than that of the denominator, so it is not necessary to do the polynomials division.

• $\dfrac{x+4}{x^2-5x+3}=\dfrac{x+4}{(x-2)(x-3)}$

• $\dfrac{x+4}{(x-2)(x-3)}=\dfrac{A}{x-2}+\dfrac{B}{x-3}$

• $\dfrac{A}{x-2}+\dfrac{B}{x-3}=\dfrac{A(x-3)+B(x-2)}{(x-2)(x-3)}$

$x=Ax+Bx$, for everything $x$, so we have the equality $1=A+B$ and also $4=-3A-2B$.

• Solving the system $\begin{array} {ll} 1=A+B \\ 4=-3A-2B \end{array}$ we have $A =-6$ and $B=5$.

• We have that $\dfrac{x+4}{x^2-5x+3}=\dfrac{-6}{x-2}+\dfrac{5}{x-3}$, and then

$$\int \frac{x+4}{x^2-5x+3} \ dx=\int\dfrac{-6}{x-2}+\dfrac{5}{x-3} \ dx=\int\dfrac{-6}{x-2} \ dx+ \int\dfrac{5}{x-3} \ dx=$$ $$=-6\cdot\ln|x-2|+5\cdot\ln|x-3|+C$$