integral(-3x^3+6x^2-10x+5)/(x^4-x^3) by using partial fraction

To solve the integral of \(\dfrac{-3x^3 + 6x^2 - 10x + 5}{x^4 - x^3}\) using partial fractions, let's go through the steps carefully.

### Step 1: Factor the denominator
We begin by factoring the denominator \(x^4 - x^3\):

\[
x^4 - x^3 = x^3(x - 1)
\]

Thus, the given expression becomes:

\[
\frac{-3x^3 + 6x^2 - 10x + 5}{x^3(x - 1)}
\]

### Step 2: Set up the partial fraction decomposition
We can express the fraction as a sum of partial fractions of the form:

\[
\frac{-3x^3 + 6x^2 - 10x + 5}{x^3(x - 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x - 1}
\]

### Step 3: Combine the right-hand side
We need to combine the partial fractions on the right-hand side:

\[
\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x - 1} = \frac{A(x^2)(x - 1) + B(x)(x - 1) + C(x - 1) + Dx^3}{x^3(x - 1)}
\]

Simplify the numerator of the right-hand side and equate it to the numerator of the original fraction \(-3x^3 + 6x^2 - 10x + 5\).

### Step 4: Solve for the constants \(A\), \(B\), \(C\), and \(D\)
Now, expand and collect like terms on the right-hand side. Once we match the coefficients of powers of \(x\), we will obtain a system of equations to solve for \(A\), \(B\), \(C\), and \(D\).

Let's expand the numerator:

\[
A(x^2)(x - 1) = A(x^3 - x^2)
\]
\[
B(x)(x - 1) = B(x^2 - x)
\]
\[
C(x - 1) = Cx - C
\]
\[
Dx^3 = Dx^3
\]

Thus, the numerator becomes:

\[
(A + D)x^3 + (B - A)x^2 + (C - B)x - C
\]

Now equate this to the original numerator \(-3x^3 + 6x^2 - 10x + 5\):

\[
(A + D)x^3 + (B - A)x^2 + (C - B)x - C = -3x^3 + 6x^2 - 10x + 5
\]

### Step 5: Set up a system of equations
We can now match the coefficients of each power of \(x\):

1. \(A + D = -3\)  (coefficient of \(x^3\))
2. \(B - A = 6\)  (coefficient of \(x^2\))
3. \(C - B = -10\)  (coefficient of \(x\))
4. \(-C = 5\)  (constant term)

### Step 6: Solve the system
From equation 4, we immediately get:

\[
C = -5
\]

Substitute \(C = -5\) into equation 3:

\[
-5 - B = -10 \implies B = 5
\]

Substitute \(B = 5\) into equation 2:

\[
5 - A = 6 \implies A = -1
\]

Substitute \(A = -1\) into equation 1:

\[
-1 + D = -3 \implies D = -2
\]

Thus, the partial fraction decomposition is:

\[
\frac{-3x^3 + 6x^2 - 10x + 5}{x^3(x - 1)} = \frac{-1}{x} + \frac{5}{x^2} + \frac{-5}{x^3} + \frac{-2}{x - 1}
\]

### Step 7: Integrate each term
Now, we integrate each term separately:

\[
\int \frac{-1}{x} \, dx = -\ln|x|
\]
\[
\int \frac{5}{x^2} \, dx = \frac{5}{x}
\]
\[
\int \frac{-5}{x^3} \, dx = \frac{5}{2x^2}
\]
\[
\int \frac{-2}{x - 1} \, dx = -2\ln|x - 1|
\]

### Step 8: Combine the results
Thus, the final result is:

\[
\int \frac{-3x^3 + 6x^2 - 10x + 5}{x^3(x - 1)} \, dx = -\ln|x| + \frac{5}{x} + \frac{5}{2x^2} - 2\ln|x - 1| + C
\]

where \(C\) is the constant of integration.

---

Would you like any details or have any further questions?

Here are 5 related questions to expand your knowledge:
1. How do we determine the form of partial fractions for more complex denominators?
2. What methods can we use if the numerator's degree is higher than the denominator's degree?
3. How do logarithmic properties apply to integrals involving fractions?
4. What are some strategies for factoring difficult polynomials in integrals?
5. Can partial fraction decomposition be used in multivariable calculus?

**Tip**: Always check if the degree of the numerator is higher than the denominator—if it is, use polynomial long division before applying partial fractions.

integral(-3x^3 + 6x^2 - 10x + 5)/(x^4 - x^3) by using partial fraction

Learn how to solve the integral of (-3x^3 + 6x^2 - 10x + 5) / (x^4 - x^3) using partial fraction decomposition. This detailed guide covers key concepts like polynomial division and the fundamental theorem of calculus, providing a step-by-step solution for college-level calculus students.

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