Let f(x)= (3x-2)/((x+1)^2(x+3))dx =, (x+1)/(x+3), +(1/(x+1))+, where is the constant of integration. If f(0)=11/4(1/3)+4, then which of the following are correct? (A) 4+2+6=22 (B) f(1)=[1- 2] (C) ++=27/4 (D) =-5/2 Choose the correct answer from the options given below: Option 1: (A), (B) and (C) only Option 2: (B), (C) and (D) only Option 3: (A) and (D) only Option 4: (B) and (C) only

Answer: (B) and (C) only. Solution: Partial fractions give (11/4)/(x+1)-(5/2)/((x+1)^2)-(11/4)/(x+3). Integrating, =11/4 and =52. Using f(0): 11/413+52+=11/413+4, so =32. Then (A) gives 2522 and (D) fails, while f(1)=11/4(1-2) and ++=27/4 hold.

Calculus

Integrals

Hard

applied

Let $f(x)=\int \frac{3x-2}{(x+1)^2(x+3)}dx = \alpha\log\left|\frac{x+1}{x+3}\right|+\beta\left(\frac{1}{x+1}\right)+\gamma$ , where $\gamma$ is the constant of integration. If $f(0)=\frac{11}{4}\log\left(\frac{1}{3}\right)+4$ , then which of the following are correct?

(A) $4\alpha+2\beta+6\gamma=22$

(B) $f(1)=\alpha[1-\log 2]$

(C) $\alpha+\beta+\gamma=\frac{27}{4}$

(D) $\gamma=-\frac{5}{2}$

Choose the correct answer from the options given below:

(A), (B) and (C) only

(B), (C) and (D) only

(A) and (D) only

(B) and (C) only

✅ Correct Option: 4

Partial fractions give $\frac{11/4}{x+1}-\frac{5/2}{(x+1)^2}-\frac{11/4}{x+3}$ .

Integrating, $\alpha=\frac{11}{4}$ and $\beta=\frac52$ . Using $f(0)$ : $\frac{11}{4}\log\frac13+\frac52+\gamma=\frac{11}{4}\log\frac13+4$ , so $\gamma=\frac32$ .

Then (A) gives $25\ne22$ and (D) fails, while $f(1)=\frac{11}{4}(1-\log2)$ and $\alpha+\beta+\gamma=\frac{27}{4}$ hold.

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