If $\int e^x\left(\frac{x-1}{(x+1)^3}\right)dx = \frac{Ae^x}{(x+1)^B} + C$, where C is constant of integration then which of the following are correct?

(A) $A = -1$

(B) $A = 1$

(C) $B = 3$

(D) $B = 2$

Choose the correct answer from the options given below:

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

For integrals involving $e^x$, the formula is:

$\int e^x[f(x) + f'(x)]dx = e^x \cdot f(x) + C$

---

From the answer form $\frac{Ae^x}{(x+1)^B}$, let:

$f(x) = \frac{A}{(x+1)^B}$

Finding $f'(x)$:

$f'(x) = A \cdot (-B) \cdot (x+1)^{-B-1}$

$f'(x) = \frac{-AB}{(x+1)^{B+1}}$

---

Calculate $f(x) + f'(x)$:

$f(x) + f'(x) = \frac{A}{(x+1)^B} + \frac{-AB}{(x+1)^{B+1}}$

$= \frac{A(x+1) - AB}{(x+1)^{B+1}}$

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

---

Matching with the given expression:

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

Comparing denominators:

$B + 1 = 3$

$B = 2$

Comparing numerators:

$A(x+1-B) = x-1$

Substituting $B = 2$:

$A(x-1) = x-1$

$A = 1$

---

Therefore:

(B) $A = 1$ is correct

(D) $B = 2$ is correct

The correct answer is: (B) and (D) only