Question

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

Answer 1

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

Explanation:

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)`