Math worksheet featuring four calculus problems on partial fractions decomposition for integration, designed for advanced high school students.

Problem 5:

Use partial fractions decomposition to find:
\[
\int \frac{1}{x^3 + 4x} \, dx
\]

#### Step 1: Factor the denominator
The denominator is \( x^3 + 4x \). Factor it:
\[
x^3 + 4x = x(x^2 + 4)
\]

#### Step 2: Set up partial fractions
We decompose the fraction:
\[
\frac{1}{x(x^2 + 4)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4}
\]

#### Step 3: Combine the fractions
Combine the right-hand side over a common denominator:
\[
\frac{A}{x} + \frac{Bx + C}{x^2 + 4} = \frac{A(x^2 + 4) + (Bx + C)x}{x(x^2 + 4)}
\]
\[
= \frac{Ax^2 + 4A + Bx^2 + Cx}{x(x^2 + 4)}
\]
\[
= \frac{(A + B)x^2 + Cx + 4A}{x(x^2 + 4)}
\]

#### Step 4: Equate coefficients
The numerator must equal 1:
\[
(A + B)x^2 + Cx + 4A = 1
\]
This gives us the system of equations:
1. \( A + B = 0 \)
2. \( C = 0 \)
3. \( 4A = 1 \)

#### Step 5: Solve the system
From \( 4A = 1 \):
\[
A = \frac{1}{4}
\]
From \( A + B = 0 \):
\[
B = -A = -\frac{1}{4}
\]
From \( C = 0 \):
\[
C = 0
\]

#### Step 6: Write the partial fractions
Substitute \( A \), \( B \), and \( C \) back:
\[
\frac{1}{x(x^2 + 4)} = \frac{\frac{1}{4}}{x} + \frac{-\frac{1}{4}x + 0}{x^2 + 4}
\]
\[
= \frac{1}{4x} - \frac{x}{4(x^2 + 4)}
\]

#### Step 7: Integrate
Integrate each term separately:
\[
\int \frac{1}{x(x^2 + 4)} \, dx = \int \left( \frac{1}{4x} - \frac{x}{4(x^2 + 4)} \right) \, dx
\]
\[
= \frac{1}{4} \int \frac{1}{x} \, dx - \frac{1}{4} \int \frac{x}{x^2 + 4} \, dx
\]

- For the first term:
\[
\frac{1}{4} \int \frac{1}{x} \, dx = \frac{1}{4} \ln |x|
\]

- For the second term, use substitution \( u = x^2 + 4 \), so \( du = 2x \, dx \) or \( \frac{du}{2} = x \, dx \):
\[
\int \frac{x}{x^2 + 4} \, dx = \int \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \ln |u| = \frac{1}{2} \ln (x^2 + 4)
\]
Thus:
\[
-\frac{1}{4} \int \frac{x}{x^2 + 4} \, dx = -\frac{1}{4} \cdot \frac{1}{2} \ln (x^2 + 4) = -\frac{1}{8} \ln (x^2 + 4)
\]

#### Step 8: Combine the results
\[
\int \frac{1}{x(x^2 + 4)} \, dx = \frac{1}{4} \ln |x| - \frac{1}{8} \ln (x^2 + 4) + C
\]

#### Final Answer:
\[
\boxed{D}
\]

---

Problem 6:

Use partial fractions decomposition to find:
\[
\int \frac{4x^2 - 7x - 17}{6x^2 - 11x - 10} \, dx
\]

#### Step 1: Perform polynomial long division
The degree of the numerator (2) is equal to the degree of the denominator (2). Perform long division:
\[
\frac{4x^2 - 7x - 17}{6x^2 - 11x - 10}
\]

Divide the leading terms:
\[
\frac{4x^2}{6x^2} = \frac{2}{3}
\]

Multiply \( \frac{2}{3} \) by the denominator:
\[
\frac{2}{3}(6x^2 - 11x - 10) = 4x^2 - \frac{22}{3}x - \frac{20}{3}
\]

Subtract this from the original numerator:
\[
(4x^2 - 7x - 17) - \left( 4x^2 - \frac{22}{3}x - \frac{20}{3} \right)
\]
\[
= 4x^2 - 7x - 17 - 4x^2 + \frac{22}{3}x + \frac{20}{3}
\]
\[
= \left( -7x + \frac{22}{3}x \right) + \left( -17 + \frac{20}{3} \right)
\]
\[
= \left( \frac{-21}{3}x + \frac{22}{3}x \right) + \left( \frac{-51}{3} + \frac{20}{3} \right)
\]
\[
= \frac{1}{3}x - \frac{31}{3}
\]

So:
\[
\frac{4x^2 - 7x - 17}{6x^2 - 11x - 10} = \frac{2}{3} + \frac{\frac{1}{3}x - \frac{31}{3}}{6x^2 - 11x - 10}
\]

#### Step 2: Factor the denominator
Factor \( 6x^2 - 11x - 10 \):
\[
6x^2 - 11x - 10 = (3x + 2)(2x - 5)
\]

#### Step 3: Decompose the remaining fraction
Decompose:
\[
\frac{\frac{1}{3}x - \frac{31}{3}}{(3x + 2)(2x - 5)} = \frac{A}{3x + 2} + \frac{B}{2x - 5}
\]

Combine:
\[
\frac{A}{3x + 2} + \frac{B}{2x - 5} = \frac{A(2x - 5) + B(3x + 2)}{(3x + 2)(2x - 5)}
\]
\[
= \frac{(2A + 3B)x + (-5A + 2B)}{(3x + 2)(2x - 5)}
\]

Equate coefficients:
\[
(2A + 3B)x + (-5A + 2B) = \frac{1}{3}x - \frac{31}{3}
\]
This gives:
1. \( 2A + 3B = \frac{1}{3} \)
2. \( -5A + 2B = -\frac{31}{3} \)

#### Step 4: Solve the system
Multiply both equations by 3 to clear the fractions:
1. \( 6A + 9B = 1 \)
2. \( -15A + 6B = -31 \)

Solve using elimination or substitution. Multiply the first equation by 2 and the second by 3:
1. \( 12A + 18B = 2 \)
2. \( -45A + 18B = -93 \)

Subtract the second from the first:
\[
(12A + 18B) - (-45A + 18B) = 2 - (-93)
\]
\[
57A = 95 \implies A = \frac{95}{57} = \frac{5}{3}
\]

Substitute \( A = \frac{5}{3} \) into \( 6A + 9B = 1 \):
\[
6\left(\frac{5}{3}\right) + 9B = 1
\]
\[
10 + 9B = 1 \implies 9B = -9 \implies B = -1
\]

#### Step 5: Write the partial fractions
\[
\frac{\frac{1}{3}x - \frac{31}{3}}{(3x + 2)(2x - 5)} = \frac{\frac{5}{3}}{3x + 2} + \frac{-1}{2x - 5}
\]

#### Step 6: Integrate
\[
\int \frac{4x^2 - 7x - 17}{6x^2 - 11x - 10} \, dx = \int \left( \frac{2}{3} + \frac{\frac{5}{3}}{3x + 2} + \frac{-1}{2x - 5} \right) \, dx
\]
\[
= \frac{2}{3} \int 1 \, dx + \frac{5}{3} \int \frac{1}{3x + 2} \, dx - \int \frac{1}{2x - 5} \, dx
\]

- For the first term:
\[
\frac{2}{3} \int 1 \, dx = \frac{2}{3}x
\]

- For the second term, use substitution \( u = 3x + 2 \), so \( du = 3 \, dx \) or \( \frac{du}{3} = dx \):
\[
\frac{5}{3} \int \frac{1}{3x + 2} \, dx = \frac{5}{3} \int \frac{1}{u} \cdot \frac{du}{3} = \frac{5}{9} \ln |u| = \frac{5}{9} \ln |3x + 2|
\]

- For the third term, use substitution \( v = 2x - 5 \), so \( dv = 2 \, dx \) or \( \frac{dv}{2} = dx \):
\[
-\int \frac{1}{2x - 5} \, dx = -\int \frac{1}{v} \cdot \frac{dv}{2} = -\frac{1}{2} \ln |v| = -\frac{1}{2} \ln |2x - 5|
\]

#### Step 7: Combine the results
\[
\int \frac{4x^2 - 7x - 17}{6x^2 - 11x - 10} \, dx = \frac{2}{3}x + \frac{5}{9} \ln |3x + 2| - \frac{1}{2} \ln |2x - 5| + C
\]

#### Final Answer:
\[
\boxed{A}
\]

---

Problem 7:

Use partial fractions decomposition to find:
\[
\int \frac{x^3 + x}{x^2 - 1} \, dx
\]

#### Step 1: Perform polynomial long division
The degree of the numerator (3) is greater than the degree of the denominator (2). Perform long division:
\[
\frac{x^3 + x}{x^2 - 1}
\]

Divide the leading terms:
\[
\frac{x^3}{x^2} = x
\]

Multiply \( x \) by the denominator:
\[
x(x^2 - 1) = x^3 - x
\]

Subtract this from the original numerator:
\[
(x^3 + x) - (x^3 - x) = x^3 + x - x^3 + x = 2x
\]

So:
\[
\frac{x^3 + x}{x^2 - 1} = x + \frac{2x}{x^2 - 1}
\]

#### Step 2: Factor the denominator
Factor \( x^2 - 1 \):
\[
x^2 - 1 = (x - 1)(x + 1)
\]

#### Step 3: Decompose the remaining fraction
Decompose:
\[
\frac{2x}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}
\]

Combine:
\[
\frac{A}{x - 1} + \frac{B}{x + 1} = \frac{A(x + 1) + B(x - 1)}{(x - 1)(x + 1)}
\]
\[
= \frac{(A + B)x + (A - B)}{(x - 1)(x + 1)}
\]

Equate coefficients:
\[
(A + B)x + (A - B) = 2x
\]
This gives:
1. \( A + B = 2 \)
2. \( A - B = 0 \)

#### Step 4: Solve the system
From \( A - B = 0 \):
\[
A = B
\]

Substitute \( A = B \) into \( A + B = 2 \):
\[
A + A = 2 \implies 2A = 2 \implies A = 1, B = 1
\]

#### Step 5: Write the partial fractions
\[
\frac{2x}{(x - 1)(x + 1)} = \frac{1}{x - 1} + \frac{1}{x + 1}
\]

#### Step 6: Integrate
\[
\int \frac{x^3 + x}{x^2 - 1} \, dx = \int \left( x + \frac{1}{x - 1} + \frac{1}{x + 1} \right) \, dx
\]
\[
= \int x \, dx + \int \frac{1}{x - 1} \, dx + \int \frac{1}{x + 1} \, dx
\]

- For the first term:
\[
\int x \, dx = \frac{x^2}{2}
\]

- For the second term:
\[
\int \frac{1}{x - 1} \, dx = \ln |x - 1|
\]

- For the third term:
\[
\int \frac{1}{x + 1} \, dx = \ln |x + 1|
\]

#### Step 7: Combine the results
\[
\int \frac{x^3 + x}{x^2 - 1} \, dx = \frac{x^2}{2} + \ln |x - 1| + \ln |x + 1| + C
\]
\[
= \frac{x^2}{2} + \ln |(x - 1)(x + 1)| + C
\]
\[
= \frac{x^2}{2} + \ln |x^2 - 1| + C
\]

#### Final Answer:
\[
\boxed{B}
\]

---

Problem 8:

Use partial fractions decomposition to find:
\[
\int \frac{x^3 - 4}{x^3 + 2x^2 + 2x} \, dx
\]

#### Step 1: Factor the denominator
The denominator is \( x^3 + 2x^2 + 2x \). Factor it:
\[
x^3 + 2x^2 + 2x = x(x^2 + 2x + 2)
\]

#### Step 2: Set up partial fractions
We decompose the fraction:
\[
\frac{x^3 - 4}{x(x^2 + 2x + 2)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 2x + 2}
\]

#### Step 3: Combine the fractions
Combine the right-hand side over a common denominator:
\[
\frac{A}{x} + \frac{Bx + C}{x^2 + 2x + 2} = \frac{A(x^2 + 2x + 2) + (Bx + C)x}{x(x^2 + 2x + 2)}
\]
\[
= \frac{Ax^2 + 2Ax + 2A + Bx^2 + Cx}{x(x^2 + 2x + 2)}
\]
\[
= \frac{(A + B)x^2 + (2A + C)x + 2A}{x(x^2 + 2x + 2)}
\]

#### Step 4: Equate coefficients
The numerator must equal \( x^3 - 4 \):
\[
(A + B)x^2 + (2A + C)x + 2A = x^3 - 4
\]
This gives us the system of equations:
1. \( A + B = 0 \)
2. \( 2A + C = 0 \)
3. \( 2A = -4 \)

#### Step 5: Solve the system
From \( 2A = -4 \):
\[
A = -2
\]
From \( A + B = 0 \):
\[
B = -A = 2
\]
From \( 2A + C = 0 \):
\[
2(-2) + C = 0 \implies -4 + C = 0 \implies C = 4
\]

#### Step 6: Write the partial fractions
Substitute \( A \), \( B \), and \( C \) back:
\[
\frac{x^3 - 4}{x(x^2 + 2x + 2)} = \frac{-2}{x} + \frac{2x + 4}{x^2 + 2x + 2}
\]

#### Step 7: Integrate
Integrate each term separately:
\[
\int \frac{x^3 - 4}{x(x^2 + 2x + 2)} \, dx = \int \left( \frac{-2}{x} + \frac{2x + 4}{x^2 + 2x + 2} \right) \, dx
\]
\[
= -2 \int \frac{1}{x} \, dx + \int \frac{2x + 4}{x^2 + 2x + 2} \, dx
\]

- For the first term:
\[
-2 \int \frac{1}{x} \, dx = -2 \ln |x|
\]

- For the second term, split it into two integrals:
\[
\int \frac{2x + 4}{x^2 + 2x + 2} \, dx = \int \frac{2x + 2}{x^2 + 2x + 2} \, dx + \int \frac{2}{x^2 + 2x + 2} \, dx
\]

- For the first part:
\[
\int \frac{2x + 2}{x^2 + 2x + 2} \, dx
\]
Use substitution \( u = x^2 + 2x + 2 \), so \( du = (2x + 2) \, dx \):
\[
\int \frac{2x + 2}{x^2 + 2x + 2} \, dx = \int \frac{1}{u} \, du = \ln |u| = \ln (x^2 + 2x + 2)
\]

- For the second part:
\[
\int \frac{2}{x^2 + 2x + 2} \, dx
\]
Complete the square in the denominator:
\[
x^2 + 2x + 2 = (x + 1)^2 + 1
\]
So:
\[
\int \frac{2}{x^2 + 2x + 2} \, dx = \int \frac{2}{(x + 1)^2 + 1} \, dx
\]
Use the arctangent formula:
\[
\int \frac{1}{(x + 1)^2 + 1} \, dx = \tan^{-1}(x + 1)
\]
Thus:
\[
\int \frac{2}{(x + 1)^2 + 1} \, dx = 2 \tan^{-1}(x + 1)
\]

#### Step 8: Combine the results
\[
\int \frac{x^3 - 4}{x(x^2 + 2x + 2)} \, dx = -2 \ln |x| + \ln (x^2 + 2x + 2) + 2 \tan^{-1}(x + 1) + C
\]

#### Final Answer:
\[
\boxed{A}
\]