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

Question

2441 views around the world

You can reuse this answer
Creative Commons License

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)#