JEE MAIN - Mathematics (2014 (Offline) - No. 6)

The integral $$\int {\left( {1 + x - {1 \over x}} \right){e^{x + {1 \over x}}}dx} $$ is equal to
JEE Main 2014 (Offline) Mathematics - Indefinite Integrals Question 73 English Option 1
JEE Main 2014 (Offline) Mathematics - Indefinite Integrals Question 73 English Option 2
JEE Main 2014 (Offline) Mathematics - Indefinite Integrals Question 73 English Option 3
JEE Main 2014 (Offline) Mathematics - Indefinite Integrals Question 73 English Option 4

Explanation

Let $$I = \int {\left( {1 + x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle x$}}}}dx$$

$$ = \int {{e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle x$}}}}} dx + \int {\left( {x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle x$}}}}dx$$

$$ = x.{e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle x$}}}} - \int {x\left( {1 - {1 \over {{x^2}}}} \right)} {e^{x+{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle x$}}}}dx$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \int {\left( {x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle x$}}}}dx$$

$$ = x.{e^{x + {1 \over x}}} - \int {\left( {x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle x$}}}}dx$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \int {\left( {x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle x$}}}}dx$$

$$ = x{e^{x + {1 \over x}}} + C$$

Comments (0)

Advertisement