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

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

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

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