Question #85687
6x-1/((x-3)(x+1)) into partial fractions
1
Expert's answer
2019-03-01T10:24:51-0500

6x1(x3)(x+1)=Ax3+Bx+1=A(x+1)+B(x3)(x3)(x+1)=Ax+A+Bx3B(x3)(x+1)=x(A+B)+A3B(x3)(x+1)\frac{6x-1}{(x-3)(x+1)}=\frac{A}{x-3}+\frac{B}{x+1}=\frac{A(x+1)+B(x-3)}{(x-3)(x+1)}=\frac{Ax+A+Bx-3B}{(x-3)(x+1)}=\frac{x(A+B)+A-3B}{(x-3)(x+1)}

{A+B=6A3B=1\begin{cases} A+B=6 \\ A-3B=-1 \end{cases}

{A+B=6A=3B1\begin{cases} A+B=6 \\ A=3B-1 \end{cases}

3B1+B=63B-1+B=6

4B=74B=7

B=74B=\frac{7}{4}

A=3741=174A=3\cdot\frac{7}{4}-1=\frac{17}{4}

6x1(x3)(x+1)=174(x3)+74(x+1)\frac{6x-1}{(x-3)(x+1)}=\frac{17}{4(x-3)}+\frac{7}{4(x+1)}


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