Evaluate the following integrals. int(dx)/(x^3-x^2)

Step 1
Consider the provided integral,
${\displaystyle \int \frac{dx}{x^3-x^2}}$
Evaluate the following integrals.
We can write as,
${\displaystyle \int \frac{dx}{x^3-x^2}=\int \frac{1}{x^2(x-1)}dx}$
First, we find the partial fraction.
${\displaystyle \frac{1}{x^2(x-1)}=\frac{a_0}{x}+\frac{a_1}{x^2}+\frac{a_3}{x-1}}$...(1)
Step 2
Now, multiply the equation by the provided denominator.
${\displaystyle \frac{1\cdot x^2(x-1)}{x^2(x-1)}=\frac{a_0{x}^2(x-1)}{x}+\frac{a_1{x}^2(x-1)}{x^2}+\frac{a_2{x}^2(x-1)}{x}-1}$
${\displaystyle 1=a_{0}x(x-1)+a_1(x-1)+a_2{x}^2}$
Compare the both the sides of the coefficient.
${\displaystyle a_1=-1,a_2=1}$
Using above values find the ${\displaystyle a_0}$.
${\displaystyle a_0=-1}$.
Step 3
Put the above values in the equation (1).
${\displaystyle \frac{1}{x^2(x-1)}=\frac{-1}{x}+\frac{-1}{x^2}+\frac{1}{x-1}}$
${\displaystyle =-\frac{1}{x}=\frac{1}{x^2}+\frac{1}{x-1}}$
So, the integrals becomes.
${\displaystyle \int \frac{dx}{x^3-x^2}=\int \frac{1}{x^2(x-1)}dx}$
${\displaystyle =-\int \frac{1}{x}dx-\int \frac{1}{x^2}dx+\int \frac{1}{x-1}dx}$
${\displaystyle =-\ln \left|x\right|-(-\frac{1}{x})+\ln \left|x-1\right|}$
${\displaystyle =-\ln \left|x\right|+\frac{1}{x}+\ln \left|x-1\right|+C}$
Hence.