To evaluate the integral

$\int \frac{x^2 - 5}{x^3 - 2x^2 + x} \, dx$

we first need to find the partial fraction decomposition of the integrand.

Step 1: Factor the denominator

The denominator of the integrand is:

$x^3 - 2x^2 + x$

We can factor out an $x$ from each term:

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

Next, observe that $x^2 - 2x + 1$ is a perfect square:

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

So the denominator becomes:

$x(x - 1)^2$

Step 2: Set up the partial fraction decomposition

The general form of the partial fraction decomposition for the expression $\frac{1}{x(x - 1)^2}$ is:

$\frac{x^2 - 5}{x(x - 1)^2} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2}$

where $A$, $B$, and $C$ are constants that we need to solve for.

Step 3: Solve for $A$, $B$, and $C$

Multiply both sides of the equation by the common denominator $x(x - 1)^2$ to eliminate the denominators:

$x^2 - 5 = A(x - 1)^2 + Bx(x - 1) + Cx$

Now expand the terms on the right-hand side:

• $A(x - 1)^2 = A(x^2 - 2x + 1) = A x^2 - 2A x + A$
• $Bx(x - 1) = B(x^2 - x) = Bx^2 - Bx$
• $Cx$

Thus, the equation becomes:

$x^2 - 5 = (A x^2 - 2A x + A) + (B x^2 - B x) + C x$

Group the terms by powers of $x$:

$x^2 - 5 = (A + B)x^2 + (-2A - B + C)x + A$

Step 4: Set up a system of equations

Now, we can equate the coefficients of powers of $x$ on both sides of the equation:

1. For $x^2$: $A + B = 1$
2. For $x$: $-2A - B + C = 0$
3. For the constant term: $A = -5$

Step 5: Solve the system

From equation 3, we know that:

$A = -5$

Substitute $A = -5$ into the first two equations:

1. $-5 + B = 1$, so $B = 6$
2. $-2(-5) - 6 + C = 0$, so $10 - 6 + C = 0$, which simplifies to $C = -4$

Step 6: Write the partial fraction decomposition

Now that we have the values of $A$, $B$, and $C$, the partial fraction decomposition is:

$\frac{x^2 - 5}{x(x - 1)^2} = \frac{-5}{x} + \frac{6}{x - 1} - \frac{4}{(x - 1)^2}$

Step 7: Rewrite the integral

Using the partial fraction decomposition, we can rewrite the integral as:

$\int \frac{x^2 - 5}{x(x - 1)^2} \, dx = \int \left( \frac{-5}{x} + \frac{6}{x - 1} - \frac{4}{(x - 1)^2} \right) \, dx$

Now the integral is split into three simpler integrals:

$\int \frac{-5}{x} \, dx + \int \frac{6}{x - 1} \, dx - \int \frac{4}{(x - 1)^2} \, dx$

Step 8: Integrate each term

1. $\int \frac{-5}{x} \, dx = -5 \ln|x|$
2. $\int \frac{6}{x - 1} \, dx = 6 \ln|x - 1|$
3. $\int \frac{-4}{(x - 1)^2} \, dx = \frac{4}{x - 1}$

Final Answer:

The integral becomes:

$-5 \ln|x| + 6 \ln|x - 1| + \frac{4}{x - 1} + C$

Where $C$ is the constant of integration.

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