COMEDK · Maths · 28. Indefinite Integration
\(\int e^x\left[\dfrac{x^2+1}{(x+1)^2}\right] d x \quad \text { is equal to }\)
- A \(e^x\left(\dfrac{x-1}{x+1}\right)+C\)
- B \(\dfrac{e^x}{x+1}+C\)
- C \(\dfrac{x e^x}{x+1}+C\)
- D \(-\dfrac{e^x}{x+1}+C\)
Answer & Solution
Correct Answer
(A) \(e^x\left(\dfrac{x-1}{x+1}\right)+C\)
Step-by-step Solution
Detailed explanation
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…
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