Q58 / 150 · MHT CET 2024 (04 May Shift 1) · Mathematics

$\int\left(1+x-\frac{1}{x}\right) \mathrm{e}^{x+\frac{1}{x}} \mathrm{~d} x$ 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
    $\quad(x-1) \mathrm{e}^{x+\frac{1}{x}}+\mathrm{c}$, (where c is a constant of integration)
  • Correct answer
    D
    $x \mathrm{e}^{x+\frac{1}{x}}+\mathrm{c}$, (where c is a constant of integration)

Solution

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

Hence, the correct answer is option: 'D'.

Relevant Questions