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MHT CET · Maths · Indefinite Integration

\(\int\left(1+x-\frac{1}{x}\right) \mathrm{e}^{x+\frac{1}{x}} \mathrm{~d} x\) 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 \(\quad(x-1) \mathrm{e}^{x+\frac{1}{x}}+\mathrm{c}\), (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

(D) \(x \mathrm{e}^{x+\frac{1}{x}}+\mathrm{c}\), (where c is a constant of integration)

Step-by-step Solution

Detailed explanation

Note that \(\int\left[x \mathrm{f}^{\prime}(x)+\mathrm{f}(x)\right] \mathrm{d} x=x \mathrm{f}(x)+\mathrm{c}\)
\(\begin{aligned}
\therefore \quad & \int\left(1+x-\frac{1}{x}\right) \mathrm{e}^{x+\frac{1}{x}} \mathrm{~d} x \\
& =\int\left[x \mathrm{e}^{x+\frac{1}{x}}\left(1-\frac{1}{x^2}\right)+\mathrm{e}^{x+\frac{1}{x}}\right] \mathrm{d} x \\
& =x \mathrm{e}^{x+\frac{1}{x}}+\mathrm{c}
\end{aligned}\)