Question

ayuda con esta integral de fracciones parciales por favor..
integral de x^2-3x+2 dx/(x+3)(x^2+1)

porfavor

Answer 1

Vemos que es una fracción propia entonces

$\dfrac{x^2-3x+2}{(x+3)(x^2+1)}=\dfrac{A}{x+3}+\dfrac{Bx+C}{x^2+1}\\ \\ \\ \dfrac{x^2-3x+2}{(x+3)(x^2+1)}=\dfrac{(A+B)x^2+(3B+C)x+(A+3C)}{(x+3)(x^2+1)}\\ \\ \\ \begin{cases} A+B=1\\ 3B+C=-3\\ A+3C=2 \end{cases}\\ \\ \\ \texttt{Resolviendo: } A=2\;,\;B=-1\;,\;C=0\\ \\ \\ \texttt{Entonces:}\\ \\ \dfrac{x^2-3x+2}{(x+3)(x^2+1)}=\dfrac{2}{x+3}-\dfrac{x}{x^2+1}$


Ahora integremos...

$\displaystyle \int\dfrac{x^2-3x+2}{(x+3)(x^2+1)}\, dx=\int\dfrac{2}{x+3}-\dfrac{x}{x^2+1}\,dx\\ \\ \\ \int\dfrac{x^2-3x+2}{(x+3)(x^2+1)}\, dx=\int\dfrac{2}{x+3}\,dx-\int\dfrac{x}{x^2+1}\,dx\\ \\ \\ \\ \boxed{\int\dfrac{x^2-3x+2}{(x+3)(x^2+1)}\, dx=2\ln|x+3|+\dfrac{1}{2}\ln (x^2+1)+K}$