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

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