How do you express $10 / [(x-1)(x^2 + 9)]$ in partial fractions?

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Answer 1 · 2016-09-23T12:59:54

$10/((x-1)(x^2+9))=1/(x-1)-(x+1)/(x^2+9)$

Let $10/((x-1)(x^2+9))hArrA/(x-1)+(Bx+C)/(x^2+9)$ .

Therefore $10/((x-1)(x^2+9))hArr(A(x^2+9)+(Bx+C)(x-1))/((x-1)(x^2+9))$

$hArr(Ax^2+9A+Bx^2-Bx+Cx-C)/((x-1)(x^2+9))$ or

$10/((x-1)(x^2+9))hArr((A+B)x^2-(B-C)x+9A-C)/((x-1)(x^2+9))$

Hence $A+B=0$ , $B-C=0$ and $9A-C=10$

or $A=-B=-C$ and so $9xx(-C)-C=10$ i.e. $-10C=10$ i.e.

$C=-1$ , $A=1$ and $B=-1$

Hence $10/((x-1)(x^2+9))=1/(x-1)-(x+1)/(x^2+9)$

How do you express 10 / [(x-1)(x^2 + 9)]10 / [(x-1)(x^2 + 9)] in partial fractions?