6:55 PM Sun 16 Mar mylab.pearson.com Kurulus Osma... Can't Open Pa... (c) CUNY Login The Evolution... Do Homework... MAT 220-S25 Rasheed Larry 03/16/25 6:56 PM $$ \equiv $$ Homework: Partial Fractions, part 2 Question 4, 8.5.33 HW Score: 40\%, 2 of 5 points Points: 0 of 1 Save Question list Evaluate $$ \int \frac{30 x^{3}-60 x^{2}+7}{x^{2}-2 x} d x $$ Question 1 $$ \int \frac{30 x^{3}-60 x^{2}+7}{x^{2}-2 x} d x=\square $$ Question 2 (Use parentheses to clearly denote the argument of each function.) Question 3 Question 4 Question 5 Help me solve this View an example Get more help • Clear all Check answer

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We begin with the integral
$$
\int \frac{30x^3-60x^2+7}{x^2-2x}\,dx.
$$

**Step 1. Perform Long Division**

Since the degree of the numerator (3) is greater than the degree of the denominator (2), we first divide $$30x^3-60x^2+7$$ by $$x^2-2x$$.

Divide the leading term:
$$
\frac{30x^3}{x^2} = 30x.
$$
Multiply the divisor by $$30x$$:
$$
30x(x^2-2x)=30x^3-60x^2.
$$
Subtract from the numerator:
$$
(30x^3-60x^2+7) - (30x^3-60x^2)=7.
$$

Thus, the division gives
$$
\frac{30x^3-60x^2+7}{x^2-2x} = 30x + \frac{7}{x^2-2x}.
$$

**Step 2. Rewrite the Integral**

The integral becomes
$$
\int \left(30x + \frac{7}{x^2-2x}\right)\,dx.
$$

Note that the denominator factors as
$$
x^2-2x=x(x-2).
$$
Thus,
$$
\int \frac{7}{x^2-2x}\,dx = \int \frac{7}{x(x-2)}\,dx.
$$

**Step 3. Partial Fractions Decomposition**

Write
$$
\frac{7}{x(x-2)}=\frac{A}{x}+\frac{B}{x-2}.
$$
Multiply both sides by $$x(x-2)$$:
$$
7 = A(x-2)+Bx.
$$
Simplify and group like terms:
$$
7 = (A+B)x -2A.
$$
For this identity to hold for all $$x$$, equate the coefficients:
$$
A+B=0,\quad -2A=7.
$$
Solve for $$A$$:
$$
-2A=7\quad\Longrightarrow\quad A=-\frac{7}{2}.
$$
Then,
$$
B=-A=\frac{7}{2}.
$$

Thus, we have
$$
\frac{7}{x(x-2)} = -\frac{7}{2}\cdot\frac{1}{x} + \frac{7}{2}\cdot\frac{1}{x-2}.
$$

**Step 4. Integrate Term-by-Term**

The original integral is now split as follows:
$$
\int 30x\,dx + \int\left( -\frac{7}{2}\cdot\frac{1}{x} + \frac{7}{2}\cdot\frac{1}{x-2} \right)dx.
$$

1. For the first term:
$$
\int 30x\,dx = 30\cdot\frac{x^2}{2}=15x^2.
$$

2. For the second term:
$$
\int -\frac{7}{2}\cdot\frac{1}{x}\,dx = -\frac{7}{2}\int \frac{1}{x}\,dx = -\frac{7}{2}\ln|x|.
$$

3. And for the third term:
$$
\int \frac{7}{2}\cdot\frac{1}{x-2}\,dx = \frac{7}{2}\int \frac{1}{x-2}\,dx = \frac{7}{2}\ln|x-2|.
$$

**Step 5. Combine the Results**

Combining all the integrated parts, we obtain
$$
15x^2 - \frac{7}{2}\ln|x| + \frac{7}{2}\ln|x-2| + C.
$$

Thus, the evaluated integral is
$$
\boxed{15x^2 - \frac{7}{2}\ln|x| + \frac{7}{2}\ln|x-2| + C}.
$$
The integral evaluates to $$15x^2 - \frac{7}{2}\ln|x| + \frac{7}{2}\ln|x-2| + C$$.

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