How do you express $\frac{x^{2} + 5 x - 7}{x^{2} {(x + 1)}^{2}}$ $(x^2 + 5x - 7 )/( x^2 (x+ 1)^2)$ in partial fractions?

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How do you express $\frac{x^{2} + 5 x - 7}{x^{2} {(x + 1)}^{2}}$ $(x^2 + 5x - 7 )/( x^2 (x+ 1)^2)$ in partial fractions?

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Answer 1 · 2016-11-04T19:19:45

The answer is $= - \frac{7}{x^{2}} + \frac{19}{x} - \frac{11}{{(x + 1)}^{2}} - \frac{19}{x + 1}$ $=-7/x^2+19/x-11/(x+1)^2-19/(x+1)$

Explanation:

$\frac{x^{2} + 5 x - 7}{x^{2} {(x + 1)}^{2}} = \frac{A}{x^{2}} + \frac{B}{x} + \frac{C}{{(x + 1)}^{2}} + \frac{D}{x + 1}$ $(x^2+5x-7)/(x^2(x+1)^2)=A/x^2+B/x+C/(x+1)^2+D/(x+1)$
$\frac{A {(x + 1)}^{2} + B x {(x + 1)}^{2} + C x^{2} + D x^{2} (x + 1)}{x^{2} {(x + 1)}^{2}}$ $(A(x+1)^2+Bx(x+1)^2+Cx^2+Dx^2(x+1))/(x^2(x+1)^2)$

$x^{2} + 5 x - 7 = A {(x + 1)}^{2} + B x {(x + 1)}^{2} + C x^{2} + D x^{2} (x + 1)$ $x^2+5x-7=A(x+1)^2+Bx(x+1)^2+Cx^2+Dx^2(x+1)$
let $x = 0$ $x=0$ $\Rightarrow$ $=>$ $- 7 = A$ $-7=A$
$x = - 1$ $x=-1$ $\Rightarrow$ $=>$ $- 11 = C$ $-11=C$
coefficentsof $x^{2}$ $x^2$ $\Rightarrow$ $=>$ $1 = A + 2 B + C + D$ $1=A+2B+C+D$
coefficients of $x$ $x$ $\Rightarrow$ $=>$ $5 = 2 A + B$ $5=2A+B$ $\Rightarrow$ $=>$ $B = 19$ $B=19$
$:.D=-19$
So, $(x^2+5x-7)/(x^2(x+1)^2)=-7/x^2+19/x-11/(x+1)^2-19/(x+1)$

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How do you express $\frac{x^{2} + 5 x - 7}{x^{2} {(x + 1)}^{2}}$ $(x^2 + 5x - 7 )/( x^2 (x+ 1)^2)$ in partial fractions?

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