Compute the integral $$\displaystyle \int \frac{x+4}{x^2-5x+3} \ dx$$
See development and solution
Development:
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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.
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$$\dfrac{x+4}{x^2-5x+3}=\dfrac{x+4}{(x-2)(x-3)}$$
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$$\dfrac{x+4}{(x-2)(x-3)}=\dfrac{A}{x-2}+\dfrac{B}{x-3}$$
- $$\dfrac{A}{x-2}+\dfrac{B}{x-3}=\dfrac{A(x-3)+B(x-2)}{(x-2)(x-3)}$$
$$x=Ax+Bx$$, for everything $$x$$, so we have the equality $$1=A+B$$ and also $$4=-3A-2B$$.
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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
$$$\int \frac{x+4}{x^2-5x+3} \ dx=\int\dfrac{-6}{x-2}+\dfrac{5}{x-3} \ dx=\int\dfrac{-6}{x-2} \ dx+ \int\dfrac{5}{x-3} \ dx=$$$ $$$=-6\cdot\ln|x-2|+5\cdot\ln|x-3|+C$$$
Solution:
$$\displaystyle \int \frac{x+4}{x^2-5x+3} \ dx= -6\cdot\ln|x-2|+5\cdot\ln|x-3|+C$$