Question

Resolve into partial fractions 9x-7/(x+3)(x²+1)

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$\frac{9x-7}{(x-3)(x^{2} +1}=\frac{-17}{5}(\frac{1}{x+3})+\frac{\frac{83}{7}(x)-\frac{6}{5}}{x^{2} +1}$

Step-by-step explanation:

       

partial fractions 9x-7/(x+3)(x²+1)

                     $\frac{9x-7}{(x+3)(x^{2} +1)} \\\frac{A}{x+3} +\frac{Bx+c}{x^{2} +1} \\now\\9x-7=A(x^{2} +1)+Bx+C(x+3)\\ sub....x=-3\\\\-34=10A\\A=\frac{-17}{5}\\ sub..x=0\\ \\-7=\frac{-17}{5}(1)+C(3)\\ \frac{-18}{5}=C(3)\\ C\frac{-6}{5}\\ sub....x=1\\2=\frac{-17}{5}(2)+(B-\frac{6}{5})(9)\\\\1= \frac{-17}{5}+(B-\frac{6}{5})(2)\\ 22=(B-\frac{6}{7})(2)\\ B=11+\frac{6}{7}\\ B=\frac{83}{7}\\ so...\frac{9x-7}{(x-3)(x^{2} +1}=\frac{-17}{5}(\frac{1}{x+3})+\frac{\frac{83}{7}(x)-\frac{6}{5}}{x^{2} +1}$

The project code is#SPJ2