Mr. X invested Rs. 4,00,000 in shares for 5 years. The value of this investment was Rs. 4,50,000 at the end of the second year, Rs. 490000 at the end…

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

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

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

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

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

Finding $f'(x)$f′(x):

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

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

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Calculate $f(x) + f'(x)$f(x)+f′(x):

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

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

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

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Matching with the given expression:

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

Comparing denominators:

$$B + 1 = 3$$B+1=3

$$B = 2$$B=2

Comparing numerators:

$$A(x+1-B) = x-1$$A(x+1−B)=x−1

Substituting $B = 2$B=2:

$$A(x-1) = x-1$$A(x−1)=x−1

$$A = 1$$A=1

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Therefore:

(B) $A = 1$A=1 is correct

(D) $B = 2$B=2 is correct

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