Let \( f(x)=\int\displaystyle\frac{3x-2}{(x+1)^2(x+3)}\,dx=\alpha\log\left|\displaystyle\frac{x+1}{x+3}\right|+\beta\left(\displaystyle\frac{1}{x+1}\right)+\gamma \), where \( \gamma \) is the constant of integration. If \( f(0)=\displaystyle\frac{11}{4}\log\left(\displaystyle\frac{1}{3}\right)+4 \), then which of the following are correct?
(A) \( 4\alpha+2\beta+6\gamma=22 \)
(B) \( f(1)=\alpha[1-\log 2] \)
(C) \( \alpha+\beta+\gamma=\displaystyle\frac{27}{4} \)
(D) \( \gamma=\displaystyle\frac{5}{2} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B) and (C) only
The integrand splits into partial fractions, with two terms for the repeated factor:
\( \displaystyle \frac{3x-2}{(x+1)^2(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+3} \)
\( \displaystyle 3x-2=A(x+1)(x+3)+B(x+3)+C(x+1)^2 \)
\( \displaystyle x=-1:\ -5=2B\ \Rightarrow\ B=-\frac{5}{2} \)
\( \displaystyle x=-3:\ -11=4C\ \Rightarrow\ C=-\frac{11}{4} \)
\( \displaystyle \text{coefficient of}\ x^2:\ A+C=0\ \Rightarrow\ A=\frac{11}{4} \)
Integrating, with \( \int\frac{dx}{(x+1)^2}=-\frac{1}{x+1} \):
\( \displaystyle f(x)=\frac{11}{4}\log|x+1|+\frac{5}{2}\cdot\frac{1}{x+1}-\frac{11}{4}\log|x+3|+\gamma \)
\( \displaystyle =\frac{11}{4}\log\left|\frac{x+1}{x+3}\right|+\frac{5}{2}\left(\frac{1}{x+1}\right)+\gamma \)
So \( \alpha=\frac{11}{4} \) and \( \beta=\frac{5}{2} \). Using f(0):
\( \displaystyle \frac{11}{4}\log\frac{1}{3}+\frac{5}{2}+\gamma=\frac{11}{4}\log\frac{1}{3}+4 \)
\( \displaystyle \gamma=\frac{3}{2} \)
At x = 1:
\( \displaystyle f(1)=\frac{11}{4}\log\frac{2}{4}+\frac{5}{2}\cdot\frac{1}{2}+\frac{3}{2} \)
\( \displaystyle =-\frac{11}{4}\log 2+\frac{11}{4} \)
\( \displaystyle =\frac{11}{4}(1-\log 2) \)
So \( \alpha=\frac{11}{4} \), \( \beta=\frac{5}{2} \), \( \gamma=\frac{3}{2} \) and \( f(1)=\frac{11}{4}(1-\log 2) \).
Checking the statements
(A) \( 4\alpha+2\beta+6\gamma=22 \) — Incorrect. \( 11+5+9=25 \), not 22.
(B) \( f(1)=\alpha[1-\log 2] \) — Correct. \( f(1)=\frac{11}{4}(1-\log 2) \) and \( \alpha=\frac{11}{4} \).
(C) \( \alpha+\beta+\gamma=\frac{27}{4} \) — Correct. \( \frac{11}{4}+\frac{10}{4}+\frac{6}{4}=\frac{27}{4} \).
(D) \( \gamma=\frac{5}{2} \) — Incorrect. \( \gamma=\frac{3}{2} \); \( \frac{5}{2} \) is the value of \( \beta \).