Integral 8dx/ (x^3-x^2)

please explain your process

integral 8/(x^3-x^2) dx = 8 (1/x+log(1-x)-log(x))+constant

you must use partial fractions.
for more info see
http://www.wolframalpha.com/input/?i=int(++8dx%2F(x^3-x^2))

1 / (x^3 - x^2) = 1 / {x^2 (x - 1) }

rewrite the numerator

1 = 1 + x^2 - x^2 + x - x

= (1 - x) + (-x^2 + x) + x^2

= -(x - 1) - x(x - 1) + x^2

so we have

1 / {x^2 (x - 1) } = { -(x - 1) - x(x - 1) + x^2} / {x^2 (x - 1)}

= - 1 / x^2 - 1 / x + 1 / (x - 1)

insert the factor of 8 in your integral, and integrate term by term to get

= 8 { 1 / x - ln | x | + ln | x - 1| } + C.............[Ans.]