TS EAMCET · Maths · Indefinite Integration
\(\int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x\) is equal to :
- A \((1+x) e^{x+x^{-1}}+C\)
- B \((x-1) e^{x+x^{-1}}+C\)
- C \(-x e^{x+x^{-1}}+C\)
- D \(x e^{x+x^{-1}}+C\)
Answer & Solution
Correct Answer
(D) \(x e^{x+x^{-1}}+C\)
Step-by-step Solution
Detailed explanation
\(\int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x\) \(=\int e^{x+x^{-1}} d x+\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x\) \(=e^{x+x^{-1}} \int d x-\int\left[\frac{d}{d x}\left(e^{x+x^{-1}}\right)\right] x d x\) \(+\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x\)…
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