AP Calculus BC 6.10 Integrating Functions Using Long Division and Completing the Square Study Notes - New Syllabus
AP Calculus BC Link Study Notes- New syllabus
AP Calculus BC Link Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Recognizing opportunities to apply knowledge of geometry and mathematical rules can simplify integration.
Key Concepts:
- Integrating Functions Using Long Division
Integrating Functions Using Long Division
Integrating Functions Using Long Division
When integrating rational functions (a polynomial divided by another polynomial), and the degree of the numerator is greater than or equal to the degree of the denominator, it is often useful to use polynomial long division before attempting other methods like substitution or partial fractions.
If you have an integral like:
\( \displaystyle \int \dfrac{P(x)}{Q(x)}\,dx \), where \( \deg(P) \geq \deg(Q) \),
perform long division first to write the expression as a polynomial plus a proper fraction:
\( \dfrac{P(x)}{Q(x)} = D(x) + \dfrac{R(x)}{Q(x)} \)
Then integrate each part separately.
Example :
Evaluate \( \displaystyle \int \dfrac{x^3 + 2x^2 + 3x + 4}{x + 1} \, dx \)
▶️Answer/Explanation
Use polynomial long division to divide \( x^3 + 2x^2 + 3x + 4 \) by \( x + 1 \).
Performing the division:
- \( x^3 \div x = x^2 \)
- Multiply: \( x^2(x + 1) = x^3 + x^2 \)
- Subtract: \( (x^3 + 2x^2) – (x^3 + x^2) = x^2 \)
- Bring down the next term: \( x^2 + 3x \)
- \( x^2 \div x = x \)
- Multiply: \( x(x + 1) = x^2 + x \)
- Subtract: \( (x^2 + 3x) – (x^2 + x) = 2x \)
- Bring down the 4: \( 2x + 4 \)
- \( 2x \div x = 2 \)
- Multiply: \( 2(x + 1) = 2x + 2 \)
- Subtract: \( (2x + 4) – (2x + 2) = 2 \)
So, \( \dfrac{x^3 + 2x^2 + 3x + 4}{x + 1} = x^2 + x + 2 + \dfrac{2}{x + 1} \)
Integrate each term:
\( \displaystyle \int \left( x^2 + x + 2 + \dfrac{2}{x + 1} \right) dx = \displaystyle \int x^2\,dx + \displaystyle \int x\,dx + \displaystyle \int 2\,dx + \displaystyle \int \dfrac{2}{x + 1}\,dx \)
= \( \dfrac{x^3}{3} + \dfrac{x^2}{2} + 2x + 2\ln|x + 1| + C \)
Example:
Evaluate \( \displaystyle \int \dfrac{x^2 + 3x + 2}{x + 1} \, dx \)
▶️Answer/Explanation
Divide \( x^2 + 3x + 2 \) by \( x + 1 \).
\( x^2 + 3x + 2 = (x + 1)(x + 2) \), so:
\( \dfrac{x^2 + 3x + 2}{x + 1} = x + 2 \)
Integrate:
\( \displaystyle \int (x + 2)\,dx = \dfrac{x^2}{2} + 2x + C \)
Example:
Evaluate \( \displaystyle \int \dfrac{2x^3 + 3x^2 – x + 5}{x^2 + 1} \, dx \)
▶️Answer/Explanation
Perform long division of \( 2x^3 + 3x^2 – x + 5 \) by \( x^2 + 1 \).
- \( 2x^3 \div x^2 = 2x \)
- Multiply: \( 2x(x^2 + 1) = 2x^3 + 2x \)
- Subtract: \( (2x^3 + 3x^2 – x + 5) – (2x^3 + 2x) = 3x^2 – 3x + 5 \)
- \( 3x^2 \div x^2 = 3 \)
- Multiply: \( 3(x^2 + 1) = 3x^2 + 3 \)
- Subtract: \( (3x^2 – 3x + 5) – (3x^2 + 3) = -3x + 2 \)
So,
\( \dfrac{2x^3 + 3x^2 – x + 5}{x^2 + 1} = 2x + 3 + \dfrac{-3x + 2}{x^2 + 1} \)
Now integrate:
\( \displaystyle \int (2x + 3)\,dx + \displaystyle \int \dfrac{-3x}{x^2 + 1}\,dx + \displaystyle \int \dfrac{2}{x^2 + 1}\,dx \)
Break it down:
- \( \displaystyle \int 2x\,dx = x^2 \)
- \( \displaystyle \int 3\,dx = 3x \)
- \( \displaystyle \int \dfrac{-3x}{x^2 + 1}\,dx = \dfrac{-3}{2} \ln(x^2 + 1) \) (by substitution)
- \( \displaystyle \int \dfrac{2}{x^2 + 1}\,dx = 2 \tan^{-1}(x) \)
Final Answer:
\( x^2 + 3x – \dfrac{3}{2} \ln(x^2 + 1) + 2 \tan^{-1}(x) + C \)
Integrating Using Completing the Square
Integrating Using Completing the Square
Sometimes, when a rational function has an irreducible quadratic expression in the denominator, completing the square allows us to express the integral in a form where standard formulas (like those involving inverse trigonometric functions) can be applied.
Step-by-step process:
- Complete the square in the denominator if needed.
- Rewrite the integral using the completed square form.
- Use substitution or inverse trigonometric identities (like \(\displaystyle \int \dfrac{1}{x^2 + a^2} dx = \dfrac{1}{a} \tan^{-1}\left( \dfrac{x}{a} \right) + C \)) to integrate.
Standard Integral Forms After Completing the Square:
| Expression | Integral |
|---|---|
| \(\displaystyle \int \dfrac{1}{x^2 + a^2} \, dx \) | \( \dfrac{1}{a} \tan^{-1}\left( \dfrac{x}{a} \right) + C \) |
| \(\displaystyle \int \dfrac{x}{x^2 + a^2} \, dx \) | \( \dfrac{1}{2} \ln(x^2 + a^2) + C \) |
Example:
Evaluate \(\displaystyle \int \dfrac{1}{x^2 + 4x + 5} \, dx \)
▶️Answer/Explanation
Step 1: Complete the square in the denominator:
\( x^2 + 4x + 5 = (x + 2)^2 + 1 \)
Step 2: Rewrite the integral:
\(\displaystyle \int \dfrac{1}{(x + 2)^2 + 1} \, dx \)
Step 3: Use substitution: Let \( u = x + 2 \Rightarrow du = dx \)
\(\displaystyle \int \dfrac{1}{u^2 + 1} \, du = \tan^{-1}(u) + C = \tan^{-1}(x + 2) + C \)
Final Answer: \( \tan^{-1}(x + 2) + C \)
Example:
Evaluate \(\displaystyle \int \dfrac{x}{x^2 + 4x + 8} \, dx \)
▶️Answer/Explanation
Step 1: Complete the square in the denominator:
\( x^2 + 4x + 8 = (x + 2)^2 + 4 \)
Step 2: Use substitution:
Let \( u = x^2 + 4x + 8 \), then \( du = (2x + 4)dx \)
Too messy. Better: Let \( u = (x + 2)^2 + 4 \), but this won’t help directly.
Better approach: Let’s use substitution \( u = x^2 + 4x + 8 \), so \( du = (2x + 4)dx \).
We can split the integral instead:
\(\displaystyle \int \dfrac{x}{x^2 + 4x + 8} dx = \dfrac{1}{2}\displaystyle \int \dfrac{2x + 4 – 4}{x^2 + 4x + 8} dx \)
\( = \dfrac{1}{2} \left(\displaystyle \int \dfrac{2x + 4}{x^2 + 4x + 8} dx -\displaystyle \int \dfrac{4}{x^2 + 4x + 8} dx \right) \)
Now:
- \(\displaystyle \int \dfrac{2x + 4}{x^2 + 4x + 8} dx = \ln|x^2 + 4x + 8| \)
- \(\displaystyle \int \dfrac{4}{(x + 2)^2 + 4} dx = 2 \tan^{-1} \left( \dfrac{x + 2}{2} \right) \)
Final Answer: \( \dfrac{1}{2} \ln|x^2 + 4x + 8| – \tan^{-1} \left( \dfrac{x + 2}{2} \right) + C \)