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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

  1. A \((x-1) \mathrm{e}^{x+\frac{1}{x}}+\mathrm{c}\), where c is a constant of integration.
  2. B \(x \mathrm{e}^{x+\frac{1}{x}}+\mathrm{c}\), where c is a constant of integration.
  3. C \((x+1) \mathrm{e}^{x+\frac{1}{x}}+\mathrm{c}\), where c is a constant of integration.
  4. D \(-x \mathrm{e}^{x+\frac{1}{x}}+\mathrm{c}\), where c is a constant of integration.
Verified Solution

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

\(\int\left(1+x-x^{-1}\right) \mathrm{e}^{x+x^{-1}} \mathrm{~d} x \)
\( =\int\left[x \mathrm{e}^{x+x^{-1}}\left(1-\frac{1}{x^2}\right)+\mathrm{e}^{x+x^{-1}}\right] \mathrm{d} x \)
\( =x \mathrm{e}^{x+x^{-1}}+\mathrm{c} \ldots\left[\because \int\left[x \mathrm{f}^{\prime}(x)+\mathrm{f}(x)\right] \mathrm{d} x=x \mathrm{f}(x)+\mathrm{c}\right]\)