How do you express # x^2/(x^2+9)^2# in partial fractions?

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Answer 1 · 2016-04-13T04:50:26

#x^2/(x^2+9)^2hArr1/(x^2+9)-9/(x^2+9)^2#

Partial fractions of #x^2/(x^2+9)^2# will be of type

#x^2/(x^2+9)^2hArr(Ax+B)/(x^2+9)+(Cx+D)/(x^2+9)^2#

Simplifying RHS

#x^2/(x^2+9)^2hArr((Ax+B)(x^2+9)+Cx+D)/(x^2+9)^2# or

#x^2/(x^2+9)^2hArr(Ax^3+Bx^2+9Ax+9B+Cx+D)/(x^2+9)^2#

Now comparing like terms on each side

#A=0#, #B=1#, #9A+C=0# and #9B+D=0#

As #A=0# and #B=1#, putting these values we get #C=0# and #D=-9#.

Hence #x^2/(x^2+9)^2hArr1/(x^2+9)-9/(x^2+9)^2#