MCQ
$\smallint \left( {1 + x - \frac{1}{x}} \right){e^{x + \frac{1}{x}}}\;dx = $
  • A
    $\;\left( {x + 1} \right){e^{x + \frac{1}{x}}}$
  • B
    $ - x{e^{x + \frac{1}{x}}}$
  • C
    $\left( {x - 1} \right){e^{x + \frac{1}{x}}}$
  • $\;x{e^{x + \frac{1}{x}}}$

Answer

Correct option: D.
$\;x{e^{x + \frac{1}{x}}}$
d
$\int\left(e^{x+\frac{1}{x}}+\left(x-\frac{1}{x}\right) e^{x+\frac{1}{x}}\right) d x$      ....$(1)$

$e^{x+\frac{1}{x}}=f(x)$

$e^{x+\frac{1}{x}}\left(1-\frac{1}{x^{2}}\right) d x=f^{\prime}(x)$

$ \Rightarrow \int {\left( {\underbrace {{e^{x + \frac{1}{x}}}}_{f\left( x \right)} + \underbrace {x \cdot {e^{x + \frac{1}{x}}}\left( {1 - \frac{1}{{{x^2}}}} \right)}_{xf'\left( x \right)}} \right)} dx$

$\Rightarrow x f(x)=x \cdot e^{x+\frac{1}{x}}+C$

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