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Use partial fraction expansion to evaluate: LaTeX: \int\frac{x-1}{x^2+3x+2}dx ∫ x − 1 x 2 + 3 x + 2 d x a. LaTeX: -2\ln\left|x+1\right|+3\ln\left|x+2\right|+C − 2 ln ⁡ | x + 1 | + 3 ln ⁡ | x + 2 | + C b. LaTeX: \frac{-2}{x+1}+\frac{3}{x+2}+C − 2 x + 1 + 3 x + 2 + C c. LaTeX: \frac{2}{\left(x+1\right)^2}+\frac{-3}{\left(x+2\right)^2}+C 2 ( x + 1 ) 2 + − 3 ( x + 2 ) 2 + C d. LaTeX: \frac{1}{\left(x+3+\frac{2}{x}\right)^2}+C 1 ( x + 3 + 2 x ) 2 + C

Respuesta :

[tex]\frac{x-1}{x^2+3x+2}\\\frac{x-1}{(x+1)(x+2)}\\\frac{x-1}{(x+1)(x+2)}=\frac{A}{x+2}+\frac{B}{x+1}\\x-1=A(x+1)+B(x+2)\\\\\text{ Let } x=-1 \text{ then } -2=A(0)+B(-1+2) \text{ which means } B=-2\\\\\text{ Instead of choose } x \text{ to be -1, let's now set it equal to -2 to find } A\\\\-3=A(-1)+B(-2+2)\\-3=-A\\A=3\\\\ \text{ So your integral becomes } \\\int \frac{3}{x+2} dx+\int \frac{-2}{x+1} dx\\3 \ln|x+2|-2\ln|x+1|+C\\\\or\\-2\ln|x+1|+3\ln|x+2|+C[/tex]

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