7:09 PM Sun 16 Mar - 35\% mylab.pearson.com Kurulus Osm... Can't Open P... CUNY Login The Evolutio... Do Homewor... Discord MAT 220-S25 Rasheed Larry 03/16/25 7:09 PM $$ \equiv $$ Homework: Partial Fractions, part 2 Question 3, 8.5.25 HW Score: 80\%, 4 of 5 points ® Points: 0 of 1 Save Question list Express the integrand as a sum of partial fractions and evaluate the integral. $$ \int \frac{24 s+24}{\left(s^{2}+1\right)(s-1)^{3}} d s $$ Question 1 Question 2 $$ \int \frac{24 s+24}{\left(s^{2}+1\right)(s-1)^{3}} d s=\square $$ Question 3 Question 4 Question 5 Help me solve this View an example Get more help Clear all Check answer

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UpStudy AI · Accepted Answer

To solve the integral

$$
\int \frac{24 s + 24}{(s^2 + 1)(s - 1)^3} \, ds,
$$

we will first express the integrand as a sum of partial fractions.

### Step 1: Set up the partial fraction decomposition

The integrand can be expressed in the form:

$$
\frac{24 s + 24}{(s^2 + 1)(s - 1)^3} = \frac{As + B}{s^2 + 1} + \frac{C}{(s - 1)} + \frac{D}{(s - 1)^2} + \frac{E}{(s - 1)^3},
$$

where $$A$$, $$B$$, $$C$$, $$D$$, and $$E$$ are constants that we need to determine.

### Step 2: Combine the right-hand side

To combine the right-hand side, we will find a common denominator:

$$
\frac{(As + B)(s - 1)^3 + C(s^2 + 1)(s - 1)^2 + D(s^2 + 1)(s - 1) + E(s^2 + 1)}{(s^2 + 1)(s - 1)^3}.
$$

### Step 3: Expand and equate coefficients

Now we will expand the numerator and set it equal to the numerator of the left-hand side:

$$
24s + 24 = (As + B)(s - 1)^3 + C(s^2 + 1)(s - 1)^2 + D(s^2 + 1)(s - 1) + E(s^2 + 1).
$$

Next, we will expand the right-hand side and collect like terms.

### Step 4: Solve for the coefficients

After expanding, we will have a polynomial in $$s$$ on both sides. We will equate the coefficients of $$s$$ and the constant terms to form a system of equations.

Let's perform the calculations to find the coefficients $$A$$, $$B$$, $$C$$, $$D$$, and $$E$$.

I'll proceed with the calculations.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}24=-A+C+D+E\0=3A-2C+2D+2E\24=B-3A+C+D+E\0=0\0=0\end{array}\right.$$
- step1: Simplify:
  $$\left\{ \begin{array}{l}24=-A+C+D+E\0=3A-2C+2D+2E\24=B-3A+C+D+E\0=0\end{array}\right.$$
- step2: Rearrange the terms:
  $$\left\{ \begin{array}{l}24=-A+C+D+E\0=3A-2C+2D+2E\24=B-3A+C+D+E\end{array}\right.$$
- step3: Solve the equation:
  $$\left\{ \begin{array}{l}A=-24+C+D+E\0=3A-2C+2D+2E\24=B-3A+C+D+E\end{array}\right.$$
- step4: Substitute the value of $$A:$$
  $$\left\{ \begin{array}{l}0=3\left(-24+C+D+E\right)-2C+2D+2E\24=B-3\left(-24+C+D+E\right)+C+D+E\end{array}\right.$$
- step5: Simplify:
  $$\left\{ \begin{array}{l}0=-72+C+5D+5E\24=B+72-2C-2D-2E\end{array}\right.$$
- step6: Solve the equation:
  $$\left\{ \begin{array}{l}0=-72+C+5D+5E\B=-48+2C+2D+2E\end{array}\right.$$
- step7: Substitute the value of $$B:$$
  $$0=-72+C+5D+5E$$
- step8: Evaluate:
  $$0=-72+5D+5E+C$$
- step9: Swap the sides:
  $$-72+5D+5E+C=0$$
- step10: Move the expression to the right side:
  $$C=0-\left(-72+5D+5E\right)$$
- step11: Subtract the terms:
  $$C=72-5D-5E$$
- step12: Substitute the value of $$C:$$
  $$B=-48+2\left(72-5D-5E\right)+2D+2E$$
- step13: Simplify:
  $$B=96-8D-8E$$
- step14: Substitute the value of $$C:$$
  $$A=-24+72-5D-5E+D+E$$
- step15: Simplify:
  $$A=48-4D-4E$$
- step16: Calculate:
  $$\left(A,B,C,D,E\right) = \left(48-4D-4E,96-8D-8E,72-5D-5E,D,E\right),\left(D,E\right) \in \mathbb{R}^{2}$$
- step17: Alternative Form:
  $$\textrm{Infinitely many solutions}$$
The system of equations derived from equating coefficients has infinitely many solutions, indicating that we need to express the partial fraction decomposition in a different way or that we have a dependency among the equations.

### Step 5: Re-evaluate the setup

Let's re-evaluate the setup of the partial fractions. The original expression is:

$$
\frac{24 s + 24}{(s^2 + 1)(s - 1)^3}.
$$

We can express it as:

$$
\frac{As + B}{s^2 + 1} + \frac{C}{(s - 1)} + \frac{D}{(s - 1)^2} + \frac{E}{(s - 1)^3}.
$$

### Step 6: Expand and collect terms again

We will expand the right-hand side again and ensure we correctly collect terms.

1. **Expand**:
   - $$(As + B)(s - 1)^3$$
   - $$C(s^2 + 1)(s - 1)^2$$
   - $$D(s^2 + 1)(s - 1)$$
   - $$E(s^2 + 1)$$

2. **Combine**:
   - Collect all terms involving $$s$$ and constant terms.

### Step 7: Set up the equations again

After expanding, we will set up the equations based on the coefficients of $$s$$ and the constant terms.

Let's perform the expansion and collection of terms again.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int \frac{24s+24}{\left(s^{2}+1\right)\left(s-1\right)^{3}} ds$$
- step1: Multiply the terms:
  $$\int \frac{24s+24}{\left(s-1\right)^{3}\left(s^{2}+1\right)} ds$$
- step2: Rewrite the fraction:
  $$\int \left(\frac{24}{\left(s-1\right)^{3}}-\frac{12}{\left(s-1\right)^{2}}+\frac{12}{s^{2}+1}\right) ds$$
- step3: Use properties of integrals:
  $$\int \frac{24}{\left(s-1\right)^{3}} ds+\int -\frac{12}{\left(s-1\right)^{2}} ds+\int \frac{12}{s^{2}+1} ds$$
- step4: Evaluate the integral:
  $$-\frac{12}{\left(s-1\right)^{2}}+\int -\frac{12}{\left(s-1\right)^{2}} ds+\int \frac{12}{s^{2}+1} ds$$
- step5: Evaluate the integral:
  $$-\frac{12}{\left(s-1\right)^{2}}+\frac{12}{s-1}+\int \frac{12}{s^{2}+1} ds$$
- step6: Evaluate the integral:
  $$-\frac{12}{\left(s-1\right)^{2}}+\frac{12}{s-1}+12\arctan\left(s\right)$$
- step7: Calculate:
  $$-\frac{12}{s^{2}-2s+1}+\frac{12}{s-1}+12\arctan\left(s\right)$$
- step8: Add the constant of integral C:
  $$-\frac{12}{s^{2}-2s+1}+\frac{12}{s-1}+12\arctan\left(s\right) + C, C \in \mathbb{R}$$
After performing the integration using partial fractions, we find that the integral evaluates to:

$$
\int \frac{24 s + 24}{(s^2 + 1)(s - 1)^3} \, ds = -\frac{12}{s^{2}-2s+1} + \frac{12}{s-1} + 12 \arctan(s) + C,
$$

where $$C$$ is the constant of integration.

### Final Result

Thus, the final answer for the integral is:

$$
-\frac{12}{(s - 1)^2} + \frac{12}{s - 1} + 12 \arctan(s) + C.
$$

You can now use this result for your homework submission. If you have any further questions or need additional help, feel free to ask!
The integral evaluates to: $$ -\frac{12}{(s - 1)^2} + \frac{12}{s - 1} + 12 \arctan(s) + C $$ where $$C$$ is the constant of integration.

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