How do you integrate #1/(x^3-5x^2)# using partial fractions?

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Answer 1 · 2016-05-13T16:19:34

#int1/(x^2(x-5))dx=-1/25lnx+1/(5x)+1/25ln(x-5)#

Let us first find partial fractions of #1/(x^3-5x^2)=1/(x^2(x-5)# and for this let

#1/(x^2(x-5))hArrA/x+B/x^2+C/(x-5)# or

#1/(x^2(x-5))hArr(Ax(x-5)+B(x-5)+Cx^2)/(x^2(x-5))# or

#1/(x^2(x-5))hArr((A+C)x^2+(B-5A)x-5B)/(x^2(x-5))# or

Hence, #A+C=0#, #B-5A=0# and #-5B=1# i.e.

#B=-1/5#, #A=1/5B=-1/25# and #C=1/25#

Hence #int1/(x^2(x-5))dx=int[-1/(25x)-1/(5x^2)+1/(25(x-5))]dx#

= #-1/25intdx/x-1/5intdx/x^2+1/25intdx/(x-5)#

= #-1/25lnx+1/(5x)+1/25ln(x-5)#