Q65 / 120 · VITEEE 2013 · Mathematics

$\int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x$ is equal to

  • A
    $(x+1) \mathrm{e}^{x+x^{-1}}+C$
  • B
    $(x-1) \mathrm{e}^{x+x^{-1}}+C$
  • Correct answer
    C
    $x e^{x+x^{-1}}+C$
  • D
    $x e^{x+x^{-1}} x+C$

Solution

$\int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x$ $=\int\left[x \cdot e^{x+x^{-1}}\left(1-\frac{1}{x^{2}}\right)+e^{x+x^{-1}}\right] d x$ $\left[\because \int x f^{\prime}(x)+f(x) d x=x f(x)+C\right]$ $\therefore \int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x=x e^{x+x^{-1}}+C$

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

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