Problems from Integrals of simple fractions

Compute the integral $$\displaystyle \int \frac{x+4}{x^2-5x+3} \ dx$$

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  • 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

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


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

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