Question

Hey MODERATORS/STARS CHALLENGE !!

Solve this if u can..
Find the correct option for the
Integration of [1+x- (1/x)]e^(x+ 1/x) dx

A) (x+1)e^(x+ 1/x) + c
B) xe^(x+ 1/x ) + c
C) (x - 1)e^(x+ 1/x) + c
D) -xe^(x+ 1/x ) + c


No Spam​

Answer 1

EXPLANATION.

$\sf \: \implies \: \int \: \bigg(1 + x - \dfrac{1}{x} \bigg) {}^{e {}^{ \bigg(x + \frac{1}{x} \bigg)} } dx$

As we know that,

$\sf \implies \:let \: we \: assume \: that \\ \\ \sf \implies \:xe {}^{ \bigg(x + \dfrac{1}{x} \bigg) } = t \\ \\ \sf \implies \: \: differentiate \: w.r.t \: \: x \: \: we \: get \\ \\ \sf \implies \: \frac{d}{dx} (uv) = u( \frac{dv}{dx} ) + v \: ( \frac{du}{dx} )$

$\sf \implies \: \: \bigg[xe {}^{ \bigg(x + \dfrac{1}{x} \bigg)} \bigg(1 - \dfrac{1}{ {x}^{2} } \bigg) \: + \: e {}^{ \bigg(x + \dfrac{1}{x} \bigg)}(1) \bigg]dx \: = dt \\ \\ \sf \: \implies \: e {}^{ \bigg(x + \dfrac{1}{x} \bigg) } \bigg[x \: - \dfrac{x}{ {x}^{2} } + 1 \bigg]dx \: = dt \\ \\ \sf \: \implies \: e {}^{ \bigg(x + \dfrac{1}{x} \bigg) } \bigg[x - \dfrac{1}{x} + 1 \bigg]dx \: = dt$

$\sf \: \implies \: e {}^{ \bigg(x + \dfrac{1}{x} \bigg) } \bigg[1 + x - \dfrac{1}{x} \bigg]dx \: = dt$

Put the value in equation, we get.

$\sf \: \implies \: \int \: dt \\ \\ \sf \: \implies \: t \: + c \\ \\ \sf \: \implies \: put \: the \: value \:of \: t \: in \: equation \: we \: get \\ \\ \sf \: \implies \: xe {}^{ \bigg(x + \dfrac{1}{x} \bigg) } + c$

Answer 2

$\rm \longrightarrow \displaystyle \int \rm \bigg(1 + x - \dfrac{1}{x} \bigg ) \large {e}^{x + \frac{1}{x} } \: dx$

$\rm \longrightarrow \displaystyle \int \rm \bigg(1 + x - \dfrac{1}{x} \bigg ) \large {e}^{x } \: {e}^{ \frac{1}{x} } \: dx$

$\rm \longrightarrow \displaystyle \int \rm \bigg( {e}^{ \frac{1}{x} } + {e}^{ \frac{1}{x} } \: x - \dfrac{ {e}^{ \frac{1}{x} } }{x} \bigg ) \large {e}^{x } \: dx$

$\rm \longrightarrow \displaystyle \int \rm \large {e}^{x } \bigg( {e}^{ \frac{1}{x} } + {e}^{ \frac{1}{x} } \: x - \dfrac{ {e}^{ \frac{1}{x} } }{x} \bigg ) \: dx$

$\rm \longrightarrow \displaystyle \int \rm \large {e}^{x } \bigg[ ({e}^{ \frac{1}{x} } \: x) + \bigg( {e}^{ \frac{1}{x} } - \dfrac{ {e}^{ \frac{1}{x} } }{x} \bigg ) \bigg] \: dx$

Now , the above expression is of the form :-

∫ eˣ[ f(x) + f'(x) ] dx

$\rightarrow \rm \dfrac{d( {e}^{ \frac{1}{x} } x)}{dx} = x \: \dfrac{d({e}^{ \frac{1}{x} })}{dx} + {e}^{ \frac{1}{x}}\dfrac{d( x)}{dx} \\ \\ \rm \: = x \: {e}^{ \frac{1}{x} } \dfrac{d({ \frac{1}{x} })}{dx} +{e}^{ \frac{1}{x}} \\ \\ \rm \: = x \: {e}^{ \frac{1}{x} } \dfrac{d({ {x}^{ - 1} })}{dx} + {e}^{ \frac{1}{x}} \\ \\ \rm \: = - x \: {e}^{ \frac{1}{x} } \dfrac{ 1}{ {x}^{2} } + {e}^{ \frac{1}{x}}\\ \\ \rm \: = - \: {e}^{ \frac{1}{x} } \dfrac{ 1}{ {x}} + {e}^{ \frac{1}{x}} \\ \\ \rm \: = {e}^{ \frac{1}{x}}- \: \dfrac{ {e}^{ \frac{1}{x} }}{ {x}}$

Now we know that :-

∫ eˣ[ f(x) + f'(x) ] dx = eˣ f(x) + K

$\rm \longrightarrow \displaystyle \int \rm \large {e}^{x } \bigg[ ({e}^{ \frac{1}{x} } \: x) + \bigg( {e}^{ \frac{1}{x} } - \dfrac{ {e}^{ \frac{1}{x} } }{x} \bigg ) \bigg] \: dx= {e}^{x }({e}^{ \frac{1}{x} } \: x) + c$

$\rm \longrightarrow \displaystyle \rm \large x \: {e}^{ (x + \frac{1}{x}) } \: + c$

Answer :

Option (B) xe^(x+ 1/x ) + c is correct