Home / AP Calculus AB 6.10 Integrating Functions Using Long Division and Completing the Square – MCQs

AP Calculus AB 6.10 Integrating Functions Using Long Division and Completing the Square - MCQs - Exam Style Questions

No-Calc Question

\(\displaystyle \int \dfrac{dx}{x^{2}-6x+10}=\ \ ?\)

(A) \(\ln|x^{2}-6x+10|+C\)
(B) \(\dfrac{1}{2x-6}\ln|x^{2}-6x+10|+C\)
(C) \(\dfrac{1}{2}\arctan(x-3)+C\)
(D) \(\arctan(x-3)+C\)

▶️ Answer/Explanation

Complete the square: \(x^{2}-6x+10=(x-3)^{2}+1\).
Use \(\displaystyle\int \dfrac{dx}{u^{2}+1}=\arctan u + C\) with \(u=x-3\).
Therefore \(\displaystyle\int\dfrac{dx}{x^{2}-6x+10}=\arctan(x-3)+C\).

Answer: (D)

No-Calc Question

\(\displaystyle \int_{1}^{2}\frac{x^{2}-x-5}{x+2}\,dx=\ ?\)

(A) \(-\dfrac{3}{2}+\ln\!\dfrac{4}{3}\)
(B) \(\dfrac{25}{2}\)
(C) \(\dfrac{5}{2}+3\ln\!\dfrac{4}{3}\)
(D) \(\dfrac{23}{45}\)

▶️ Answer/Explanation

Divide: \(\dfrac{x^{2}-x-5}{x+2}=x-3+\dfrac{1}{x+2}\). Then
\(\displaystyle \int_{1}^{2}\!\left(x-3+\frac{1}{x+2}\right)\!dx=\Big(\tfrac{x^{2}}{2}-3x+\ln|x+2|\Big)\Big|_{1}^{2}=-\tfrac{3}{2}+\ln\!\tfrac{4}{3}\).
Answer: (A)

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