If ∫ e^x ≤ft((x-1)/((x+1)^3)) dx =(A. e^x)/((x + 1)^B)+C, where C is…

The correct answer is Option (3) → (B) and (D) only

(A) A = -1 (Incorrect)
(B) A = 1 (Correct)
(C) B = 3 (Incorrect)
(D) B = 2 (Correct)

Given:

$\int e^x \left( \frac{x - 1}{(x + 1)^3} \right) dx = \frac{A e^x}{(x + 1)^B} + C$

Differentiate RHS:

$\frac{d}{dx} \left( \frac{A e^x}{(x + 1)^B} \right)$

$= A \cdot \left( \frac{e^x}{(x + 1)^B} - \frac{B e^x}{(x + 1)^{B + 1}} \right)$

$= A e^x \left( \frac{1}{(x + 1)^B} - \frac{B}{(x + 1)^{B + 1}} \right)$

Compare with the integrand: $e^x \left( \frac{x - 1}{(x + 1)^3} \right)$

Cancel $e^x$ from both sides:

$\frac{x - 1}{(x + 1)^3} = A \left( \frac{1}{(x + 1)^B} - \frac{B}{(x + 1)^{B + 1}} \right)$

Assume $B = 3$

Then RHS becomes:

$A \left( \frac{1}{(x + 1)^3} - \frac{3}{(x + 1)^4} \right)$

Make common denominator:

$= A \cdot \frac{(x + 1) - 3}{(x + 1)^4} = A \cdot \frac{x - 2}{(x + 1)^4}$

≠ LHS

Now assume $B = 2$

RHS becomes:

$A \left( \frac{1}{(x + 1)^2} - \frac{2}{(x + 1)^3} \right)$

Common denominator:

$= A \cdot \frac{(x + 1) - 2}{(x + 1)^3} = A \cdot \frac{x - 1}{(x + 1)^3}$

Matches LHS

⇒ $A = 1$, $B = 2$

Correct options: (B) A = 1 and (D) B = 2