Q69 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
$\int e^x\left[\dfrac{x^2+1}{(x+1)^2}\right] d x \quad \text { is equal to }$
- Correct answerA$e^x\left(\dfrac{x-1}{x+1}\right)+C$
- B$\dfrac{e^x}{x+1}+C$
- C$\dfrac{x e^x}{x+1}+C$
- D$-\dfrac{e^x}{x+1}+C$
Solution
The integral is of the form $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$.
We rewrite the integrand as follows:
$\dfrac{x^2 + 1}{(x + 1)^2} = \dfrac{x^2 - 1 + 2}{(x + 1)^2} = \dfrac{(x - 1)(x + 1) + 2}{(x + 1)^2} = \dfrac{x - 1}{x + 1} + \dfrac{2}{(x + 1)^2}$
Let $f(x) = \dfrac{x - 1}{x + 1}$.
Then $f'(x) = \dfrac{(x + 1)(1) - (x - 1)(1)}{(x + 1)^2} = \dfrac{x + 1 - x + 1}{(x + 1)^2} = \dfrac{2}{(x + 1)^2}$.
Since the integrand is $e^x [f(x) + f'(x)]$, the integral is $e^x f(x) + C$.
Therefore, the integral is $e^x \left( \dfrac{x - 1}{x + 1} \right) + C$.
Answer: $e^x\left(\dfrac{x-1}{x+1}\right)+C$
We rewrite the integrand as follows:
$\dfrac{x^2 + 1}{(x + 1)^2} = \dfrac{x^2 - 1 + 2}{(x + 1)^2} = \dfrac{(x - 1)(x + 1) + 2}{(x + 1)^2} = \dfrac{x - 1}{x + 1} + \dfrac{2}{(x + 1)^2}$
Let $f(x) = \dfrac{x - 1}{x + 1}$.
Then $f'(x) = \dfrac{(x + 1)(1) - (x - 1)(1)}{(x + 1)^2} = \dfrac{x + 1 - x + 1}{(x + 1)^2} = \dfrac{2}{(x + 1)^2}$.
Since the integrand is $e^x [f(x) + f'(x)]$, the integral is $e^x f(x) + C$.
Therefore, the integral is $e^x \left( \dfrac{x - 1}{x + 1} \right) + C$.
Answer: $e^x\left(\dfrac{x-1}{x+1}\right)+C$
Hence, the correct answer is option: 'A'.