AP Calculus BC 6.12 Integrating Using Linear Partial Fractions - Exam Style Questions - MCQs - New Syllabus
▶️ Answer/Explanation
Detailed solution
Use partial fractions: \( \dfrac{4}{(x+3)(x-1)}=\dfrac{1}{x-1}-\dfrac{1}{x+3} \).
\( \displaystyle \int_{2}^{3}\!\Big(\dfrac{1}{x-1}-\dfrac{1}{x+3}\Big)\,dx = \Big[\ln|x-1|-\ln|x+3|\Big]_{2}^{3} = \ln\!\big(\tfrac{5}{3}\big). \)
✅ Correct: (B)
No-CalcQuestion
Evaluate \( \displaystyle \int \frac{1}{x^{2}-4x+3}\,dx \).
(A) \( \ln|x^{2}-4x+3|+C \)
(B) \( \ln|x-3|-\ln|x-1|+C \)
(C) \( \dfrac{1}{2}\ln|x-3|-\dfrac{1}{2}\ln|x-1|+C \)
(D) \( 2\ln|x-3|-2\ln|x-1|+C \)
(B) \( \ln|x-3|-\ln|x-1|+C \)
(C) \( \dfrac{1}{2}\ln|x-3|-\dfrac{1}{2}\ln|x-1|+C \)
(D) \( 2\ln|x-3|-2\ln|x-1|+C \)
▶️ Answer/Explanation
Detailed solution
Factor the quadratic: \( x^{2}-4x+3=(x-1)(x-3) \).
Partial fractions: \( \displaystyle \frac{1}{(x-1)(x-3)}=\frac{A}{x-1}+\frac{B}{x-3} \).
Solve \( 1=A(x-3)+B(x-1) \): set \(x=3\Rightarrow B=\tfrac12\); set \(x=1\Rightarrow A=-\tfrac12\).
Hence \[ \int \frac{1}{x^{2}-4x+3}\,dx = -\frac12\ln|x-1|+\frac12\ln|x-3|+C = \frac12\ln|x-3|-\frac12\ln|x-1|+C. \] ✅ Answer: (C)
Factor the quadratic: \( x^{2}-4x+3=(x-1)(x-3) \).
Partial fractions: \( \displaystyle \frac{1}{(x-1)(x-3)}=\frac{A}{x-1}+\frac{B}{x-3} \).
Solve \( 1=A(x-3)+B(x-1) \): set \(x=3\Rightarrow B=\tfrac12\); set \(x=1\Rightarrow A=-\tfrac12\).
Hence \[ \int \frac{1}{x^{2}-4x+3}\,dx = -\frac12\ln|x-1|+\frac12\ln|x-3|+C = \frac12\ln|x-3|-\frac12\ln|x-1|+C. \] ✅ Answer: (C)