The integral provided is:

$\int \frac{dx}{x^3 - x^2}.$

Step-by-Step Solution:

Step 1: Factor the denominator

The denominator $x^3 - x^2$ can be factored as: $x^3 - x^2 = x^2(x - 1).$

Thus, the integral becomes: $\int \frac{dx}{x^2(x - 1)}.$

Step 2: Decompose into partial fractions

We express: $\frac{1}{x^2(x - 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 1},$ where $A$, $B$, and $C$ are constants to be determined.

Multiply through by $x^2(x - 1)$ to get: $1 = A x (x - 1) + B (x - 1) + C x^2.$

Expanding: $1 = A x^2 - A x + B x - B + C x^2.$

Combine like terms: $1 = (A + C)x^2 + (B - A)x - B.$

Now, equate coefficients for $x^2$, $x$, and the constant term:

• Coefficient of $x^2$: $A + C = 0$,
• Coefficient of $x$: $B - A = 0$,
• Constant term: $-B = 1$.

From $-B = 1$, we get $B = -1$.

From $B - A = 0$, we get $A = B = -1$.

From $A + C = 0$, we get $C = -A = 1$.

Thus: $\frac{1}{x^2(x - 1)} = \frac{-1}{x} - \frac{1}{x^2} + \frac{1}{x - 1}.$

Step 3: Integrate each term

Now, integrate term by term: $\int \frac{dx}{x^2(x - 1)} = \int \frac{-1}{x} \, dx - \int \frac{1}{x^2} \, dx + \int \frac{1}{x - 1} \, dx.$

The integrals are:

• $\int \frac{-1}{x} \, dx = -\ln|x|$,
• $\int \frac{1}{x^2} \, dx = -\frac{1}{x}$,
• $\int \frac{1}{x - 1} \, dx = \ln|x - 1|$.

Combine the results: $\int \frac{dx}{x^2(x - 1)} = -\ln|x| + \frac{1}{x} + \ln|x - 1| + C,$ where $C$ is the constant of integration.

Final Answer:

$\int \frac{dx}{x^2(x - 1)} = -\ln|x| + \frac{1}{x} + \ln|x - 1| + C.$

Let me know if you have questions or need further clarification!

Related Questions:

1. How do you determine partial fractions for higher powers in the denominator?
2. What happens if the integrand is non-factorable?
3. How do you integrate rational functions with repeated linear factors?
4. Why is the natural logarithm used in integrals like this one?
5. How do you check if the decomposition of partial fractions is correct?

Tip:

Always check if the denominator can be factored before attempting partial fractions. It simplifies the integration significantly.