SAT Prep Assignment: Integration using Partial Fractions - 18 practice problems for evaluating indefinite integrals.

Problem: Evaluate the following indefinite integrals using partial fractions.

We will solve each integral step by step. Here is the plan for each integral:
1. Factorize the denominator if possible.
2. Decompose the fraction into partial fractions.
3. Integrate each term separately.

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#### Integral (1):
\[
\int \frac{1}{2x^3 + x^2 - x} \, dx
\]

Step 1: Factorize the denominator.
\[
2x^3 + x^2 - x = x(2x^2 + x - 1)
\]
Factorize \(2x^2 + x - 1\):
\[
2x^2 + x - 1 = (2x - 1)(x + 1)
\]
So, the denominator becomes:
\[
2x^3 + x^2 - x = x(2x - 1)(x + 1)
\]

Step 2: Decompose into partial fractions.
\[
\frac{1}{x(2x - 1)(x + 1)} = \frac{A}{x} + \frac{B}{2x - 1} + \frac{C}{x + 1}
\]
Multiply through by the denominator \(x(2x - 1)(x + 1)\):
\[
1 = A(2x - 1)(x + 1) + Bx(x + 1) + Cx(2x - 1)
\]
Expand and collect like terms:
\[
1 = A(2x^2 + x - x - 1) + B(x^2 + x) + C(2x^2 - x)
\]
\[
1 = A(2x^2 - 1) + B(x^2 + x) + C(2x^2 - x)
\]
\[
1 = (2A + B + 2C)x^2 + (A + B - C)x + (-A)
\]
Equate coefficients:
\[
\begin{cases}
2A + B + 2C = 0 \\
A + B - C = 0 \\
-A = 1
\end{cases}
\]
From \(-A = 1\), we get \(A = -1\).
Substitute \(A = -1\) into the other equations:
\[
2(-1) + B + 2C = 0 \implies -2 + B + 2C = 0 \implies B + 2C = 2 \quad \text{(1)}
\]
\[
-1 + B - C = 0 \implies B - C = 1 \quad \text{(2)}
\]
Solve the system of linear equations (1) and (2):
From (2): \(B = C + 1\).
Substitute \(B = C + 1\) into (1):
\[
(C + 1) + 2C = 2 \implies 3C + 1 = 2 \implies 3C = 1 \implies C = \frac{1}{3}
\]
Then, \(B = C + 1 = \frac{1}{3} + 1 = \frac{4}{3}\).

So, the partial fractions are:
\[
\frac{1}{x(2x - 1)(x + 1)} = \frac{-1}{x} + \frac{\frac{4}{3}}{2x - 1} + \frac{\frac{1}{3}}{x + 1}
\]

Step 3: Integrate each term.
\[
\int \frac{1}{x(2x - 1)(x + 1)} \, dx = \int \left( \frac{-1}{x} + \frac{\frac{4}{3}}{2x - 1} + \frac{\frac{1}{3}}{x + 1} \right) \, dx
\]
\[
= -\ln|x| + \frac{4}{3} \cdot \frac{1}{2} \ln|2x - 1| + \frac{1}{3} \ln|x + 1| + C
\]
\[
= -\ln|x| + \frac{2}{3} \ln|2x - 1| + \frac{1}{3} \ln|x + 1| + C
\]

Final Answer:
\[
\boxed{-\ln|x| + \frac{2}{3} \ln|2x - 1| + \frac{1}{3} \ln|x + 1| + C}
\]

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#### Integral (2):
\[
\int \frac{3x^3 - 5x^2 - 11x + 9}{x^2 - 2x - 3} \, dx
\]

Step 1: Perform polynomial long division.
Divide \(3x^3 - 5x^2 - 11x + 9\) by \(x^2 - 2x - 3\):
\[
3x^3 - 5x^2 - 11x + 9 = (x^2 - 2x - 3)(3x + 1) + (-8x + 12)
\]
So,
\[
\frac{3x^3 - 5x^2 - 11x + 9}{x^2 - 2x - 3} = 3x + 1 + \frac{-8x + 12}{x^2 - 2x - 3}
\]

Step 2: Factorize the denominator of the remainder.
\[
x^2 - 2x - 3 = (x - 3)(x + 1)
\]
So,
\[
\frac{-8x + 12}{x^2 - 2x - 3} = \frac{-8x + 12}{(x - 3)(x + 1)}
\]
Decompose into partial fractions:
\[
\frac{-8x + 12}{(x - 3)(x + 1)} = \frac{A}{x - 3} + \frac{B}{x + 1}
\]
Multiply through by \((x - 3)(x + 1)\):
\[
-8x + 12 = A(x + 1) + B(x - 3)
\]
Expand and collect like terms:
\[
-8x + 12 = Ax + A + Bx - 3B
\]
\[
-8x + 12 = (A + B)x + (A - 3B)
\]
Equate coefficients:
\[
\begin{cases}
A + B = -8 \\
A - 3B = 12
\end{cases}
\]
Solve the system of linear equations:
From \(A + B = -8\), we get \(A = -8 - B\).
Substitute \(A = -8 - B\) into \(A - 3B = 12\):
\[
-8 - B - 3B = 12 \implies -8 - 4B = 12 \implies -4B = 20 \implies B = -5
\]
Then, \(A = -8 - (-5) = -3\).

So, the partial fractions are:
\[
\frac{-8x + 12}{(x - 3)(x + 1)} = \frac{-3}{x - 3} + \frac{-5}{x + 1}
\]

Step 3: Integrate each term.
\[
\int \frac{3x^3 - 5x^2 - 11x + 9}{x^2 - 2x - 3} \, dx = \int (3x + 1) \, dx + \int \left( \frac{-3}{x - 3} + \frac{-5}{x + 1} \right) \, dx
\]
\[
= \frac{3x^2}{2} + x - 3 \ln|x - 3| - 5 \ln|x + 1| + C
\]

Final Answer:
\[
\boxed{\frac{3x^2}{2} + x - 3 \ln|x - 3| - 5 \ln|x + 1| + C}
\]

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Continue solving the remaining integrals similarly. Each integral follows a similar process of factorization, decomposition into partial fractions, and integration. If you need solutions for specific integrals, let me know!