integeral (-4r^2)/(r+4) use partial fractions

To evaluate the integral of \(\frac{-4r^2}{r+4}\) using partial fractions, we first note that the degree of the numerator is higher than the degree of the denominator, which suggests that we should perform polynomial division before applying partial fractions.

### Step 1: Polynomial Division

We want to divide \(-4r^2\) by \(r + 4\). Let's do this division.

\[
\frac{-4r^2}{r + 4}
\]

Performing the division:

- Divide the first term of the numerator by the first term of the denominator: \(\frac{-4r^2}{r} = -4r\).
- Multiply \(-4r\) by \(r + 4\): \(-4r(r + 4) = -4r^2 - 16r\).
- Subtract this from the original numerator:
  
  \[
  (-4r^2) - (-4r^2 - 16r) = 16r.
  \]

- Now divide \(16r\) by \(r + 4\):
  
  \[
  \frac{16r}{r+4}
  \]

This fraction can no longer be divided further, so we now express the original function as:

\[
\frac{-4r^2}{r+4} = -4r + \frac{16r}{r+4}
\]

### Step 2: Partial Fraction Decomposition

Next, we decompose \(\frac{16r}{r+4}\). This expression is already in its simplest form and doesn't need further decomposition since the denominator is linear.

Thus, the integral becomes:

\[
\int \left( -4r + \frac{16r}{r+4} \right) \, dr
\]

### Step 3: Integration

Now, integrate each term separately:

1. \(\int -4r \, dr = -2r^2\)
2. \(\int \frac{16r}{r+4} \, dr\)

To integrate \(\frac{16r}{r+4}\), perform a substitution. Let \(u = r + 4\), so that \(du = dr\) and \(r = u - 4\). The integral becomes:

\[
\int \frac{16(u - 4)}{u} \, du = \int 16 \left( 1 - \frac{4}{u} \right) du
\]

This simplifies to:

\[
16 \int 1 \, du - 16 \int \frac{4}{u} \, du = 16u - 64 \ln |u|
\]

Substitute back \(u = r + 4\):

\[
16(r + 4) - 64 \ln |r + 4| = 16r + 64 - 64 \ln |r + 4|
\]

### Step 4: Final Solution

Now, putting everything together:

\[
\int \frac{-4r^2}{r+4} \, dr = -2r^2 + 16r + 64 - 64 \ln |r + 4| + C
\]

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

### Summary:

The integral of \(\frac{-4r^2}{r+4}\) is:

\[
\boxed{-2r^2 + 16r + 64 - 64 \ln |r + 4| + C}
\]

Would you like more details on any specific step? Here are 5 questions to expand your understanding:

1. Why is polynomial division necessary before using partial fractions in this case?
2. Can we apply partial fractions directly without polynomial division?
3. How does substitution simplify the integral of the second term?
4. What are the key conditions to use partial fractions effectively?
5. What are the alternative methods for integrating rational functions?

**Tip:** Always check the degree of the numerator and denominator in rational functions to determine whether polynomial division is required.

This page provides a detailed solution for evaluating the integral of (-4r^2)/(r+4) using partial fractions. Learn about polynomial division, substitution methods, and partial fraction decomposition in calculus. Ideal for undergraduate calculus students.

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