Q84 / 180 · COMEDK 2025 (25 May Shift) · Mathematics
$\int \dfrac{(x-1)}{(x+1)^3} e^x d x=P(x)+c$ then $P(x)=$
- A$-\dfrac{e^x}{x+1}$
- B$\dfrac{e^x}{x+1}$
- C$\dfrac{e^x}{(x+1)^3}$
- Correct answerD$\dfrac{e^x}{(x+1)^2}$
Solution
The integral is given by $I = \int \dfrac{x-1}{(x+1)^3} e^x dx$.
Rewrite the numerator $x-1$ as $(x+1) - 2$:
$I = \int \dfrac{(x+1) - 2}{(x+1)^3} e^x dx = \int \left( \dfrac{x+1}{(x+1)^3} - \dfrac{2}{(x+1)^3} \right) e^x dx$
$I = \int \left( \dfrac{1}{(x+1)^2} - \dfrac{2}{(x+1)^3} \right) e^x dx$
Recall the standard integral form $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$.
Let $f(x) = \dfrac{1}{(x+1)^2}$. Then $f'(x) = -2(x+1)^{-3} = -\dfrac{2}{(x+1)^3}$.
Substituting these into the integral formula:
$I = e^x \left( \dfrac{1}{(x+1)^2} \right) + c = \dfrac{e^x}{(x+1)^2} + c$.
Thus, $P(x) = \dfrac{e^x}{(x+1)^2}$.
Answer: $\dfrac{e^x}{(x+1)^2}$
Rewrite the numerator $x-1$ as $(x+1) - 2$:
$I = \int \dfrac{(x+1) - 2}{(x+1)^3} e^x dx = \int \left( \dfrac{x+1}{(x+1)^3} - \dfrac{2}{(x+1)^3} \right) e^x dx$
$I = \int \left( \dfrac{1}{(x+1)^2} - \dfrac{2}{(x+1)^3} \right) e^x dx$
Recall the standard integral form $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$.
Let $f(x) = \dfrac{1}{(x+1)^2}$. Then $f'(x) = -2(x+1)^{-3} = -\dfrac{2}{(x+1)^3}$.
Substituting these into the integral formula:
$I = e^x \left( \dfrac{1}{(x+1)^2} \right) + c = \dfrac{e^x}{(x+1)^2} + c$.
Thus, $P(x) = \dfrac{e^x}{(x+1)^2}$.
Answer: $\dfrac{e^x}{(x+1)^2}$
Hence, the correct answer is option: 'D'.