Question 372 of 416MediumMCQ
The integral \[\int {\left( {1 + x - {1 \over x}} \right){e^{x + {1 \over x}}}dx} \] is equal to
Solutions · 1
Official solution
Let \[I = \int {\left( {1 + x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{1}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{x}}}}dx\]
\[ = \int {{e^{x + {\raise0.5ex\hbox{1}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{x}}}}} dx + \int {\left( {x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{1}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{x}}}}dx\]
\[ = x.{e^{x + {\raise0.5ex\hbox{1}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{x}}}} - \int {x\left( {1 - {1 \over {{x^2}}}} \right)} {e^{x+{\raise0.5ex\hbox{1}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{x}}}}dx\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \int {\left( {x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{1}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{x}}}}dx\]
\[ = x.{e^{x + {1 \over x}}} - \int {\left( {x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{1}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{x}}}}dx\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \int {\left( {x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{1}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{x}}}}dx\]
\[ = x{e^{x + {1 \over x}}} + C\]
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