Question

Integral por fracões parciais \frac{x+1}{x^3+x^2-6x}

Answer 1

$\int {\dfrac{x+1}{x^3+x^2-6x}} \, dx \\\\ \\\int {\dfrac{x+1}{x(x+3)(x-2)}} \, dx \\\\\\\dfrac{x+1}{x(x+3)(x-2)}=\dfrac{A}{x}+\dfrac{B}{x+3}+\dfrac{C}{x-2}\\\\\\Ax^2+Ax-6A+Bx^2-2Bx+Cx^2+3Cx=x+1\\\\(A+B+C)x^2+(A-2B+3C)x+(-6A)=0x^2+1x+1$

$\left\{\begin{matrix}A+B+C=0\\A-2B+3C=1 \\ -6A=1\end{matrix}\right.\\\\\\A=-\dfrac16,\ B=-\dfrac{2}{15}},\ C=\dfrac{3}{10}\\\\\\\dfrac{x+1}{x^3+x^2-6x}=-\dfrac16\cdot\dfrac1x-\dfrac{2}{15}\cdot\dfrac1{x+3}+\dfrac3{10}\cdot\dfrac1{x-2}\\\\\\ \int {\dfrac{x+1}{x^3+x^2-6x}} \, dx =-\dfrac16 \int {\dfrac1x} \, dx -\dfrac{2}{15} \int {\dfrac{1}{x+3}} \, dx +\dfrac{3}{10}\int {\dfrac1{x-2}} \, dx= \\\\\\\boxed{-\dfrac16\ln|x|-\dfrac{2}{15}\ln|x+3|+\dfrac{3}{10}\ln|x-2|+C}$