MHT CET · Maths · Indefinite Integration
\(\int\left(1+x-\frac{1}{x}\right) \mathrm{e}^{x+\frac{1}{x}} \mathrm{~d} x\) is equal to
- A \((x-1) \mathrm{e}^{x+\frac{1}{x}}+\mathrm{c}\), where c is a constant of integration.
- B \(x \mathrm{e}^{x+\frac{1}{x}}+\mathrm{c}\), where c is a constant of integration.
- C \((x+1) \mathrm{e}^{x+\frac{1}{x}}+\mathrm{c}\), where c is a constant of integration.
- D \(-x \mathrm{e}^{x+\frac{1}{x}}+\mathrm{c}\), where c is a constant of integration.
Answer & Solution
Correct Answer
(B) \(x \mathrm{e}^{x+\frac{1}{x}}+\mathrm{c}\), where c is a constant of integration.
Step-by-step Solution
Detailed explanation
\(\int\left(1+x-x^{-1}\right) \mathrm{e}^{x+x^{-1}} \mathrm{~d} x \)
\( =\int\left[x \mathrm{e}^{x+x^{-1}}\left(1-\frac{1}{x^2}\right)+\mathrm{e}^{x+x^{-1}}\right] \mathrm{d} x \)
\( =x \mathrm{e}^{x+x^{-1}}+\mathrm{c} \ldots\left[\because \int\left[x \mathrm{f}^{\prime}(x)+\mathrm{f}(x)\right] \mathrm{d} x=x \mathrm{f}(x)+\mathrm{c}\right]\)
\( =\int\left[x \mathrm{e}^{x+x^{-1}}\left(1-\frac{1}{x^2}\right)+\mathrm{e}^{x+x^{-1}}\right] \mathrm{d} x \)
\( =x \mathrm{e}^{x+x^{-1}}+\mathrm{c} \ldots\left[\because \int\left[x \mathrm{f}^{\prime}(x)+\mathrm{f}(x)\right] \mathrm{d} x=x \mathrm{f}(x)+\mathrm{c}\right]\)
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