JEE Main 2019 — Mathematics Calculus

Let $\displaystyle {I}_{1}={\int }_{1}^{e}{(\frac{x}{e})}^{2x}{{\mathrm{log}}_{e} x}_{ }⁡dx$I1​=∫1e​(ex​)2xloge​x​⁡dx

Let ${(\frac{x}{e})}^{x}=t$(ex​)x=t

Taking natural logarithm on both sides, we get

$$x \text{ln}\frac{x}{e} =\text{ln }t$$xlnex​=ln t

$$\Rightarrow x(\text{ln }x-1)=\text{ln} t.........(i)$$⇒x(ln x−1)=lnt.........(i)

$x$x	$t$t
$1$1	$1(\mathrm{ln}1-1)=\mathrm{ln}t\Rightarrow t=\frac{1}{e}$1(ln1−1)=lnt⇒t=e1​
$e$e	$e(\mathrm{ln}e-1)=\mathrm{ln}t\Rightarrow t=1$e(lne−1)=lnt⇒t=1

Differentiating equation $(i)$(i) on both the sides, we get

$$\text{ln} x-1+x(\frac{1}{x})dx=\frac{1}{t}dt$$lnx−1+x(x1​)dx=t1​dt

$$\text{⇒ln} x dx=\frac{1}{t}dt$$⇒lnxdx=t1​dt

$$\therefore {I}_{1}={\int }_{\frac{1}{e}}^{1}{t}^{2}\frac{1}{t}dt={(\frac{{t}^{2}}{2})}_{\frac{1}{e}}^{1}=\frac{1}{2}-\frac{1}{2{e}^{2}}$$∴I1​=∫e1​1​t2t1​dt=(2t2​)e1​1​=21​−2e21​

and $\displaystyle {I}_{2}={\int }_{1}^{e}{(\frac{e}{x})}^{x}{\mathrm{log}}_{e} x⁡dx$I2​=∫1e​(xe​)xloge​x⁡dx

Let ${(\frac{e}{x})}^{x}=t$(xe​)x=t

Taking natural logarithm on both sides, we get

$$x\mathrm{ln}(\frac{e}{x})=\mathrm{ln}t$$xln(xe​)=lnt

$$\Rightarrow x(1-\mathrm{ln}x)=\mathrm{ln}t.........(ii)$$⇒x(1−lnx)=lnt.........(ii)

$x$x	$t$t
$1$1	$1(1-\mathrm{ln}1)=\mathrm{ln}t\Rightarrow t=e$1(1−ln1)=lnt⇒t=e
$e$e	$e(1-\mathrm{ln}e)=\mathrm{ln}t\Rightarrow t=1$e(1−lne)=lnt⇒t=1

Differentiating equation $(ii)$(ii) both sides, we get

$\Rightarrow [x(0-\frac{1}{x})+(1-\mathrm{ln}x).1]dx=\frac{dt}{t}$⇒[x(0−x1​)+(1−lnx).1]dx=tdt​ (using product rule of differentiation)

$$\Rightarrow \mathrm{ln}x=-\frac{dt}{t}$$⇒lnx=−tdt​

$$\therefore {I}_{2}={\int }_{e}^{1}t.\frac{-dt}{t}={(-t)}_{e}^{1}=(-1)-(-e)=e-1$$∴I2​=∫e1​t.t−dt​=(−t)e1​=(−1)−(−e)=e−1

Hence, required integral is ${I}_{1}-{I}_{2}$I1​−I2​

$$=(\frac{1}{2}-\frac{1}{2{e}^{2}})-(e-1)$$=(21​−2e21​)−(e−1)

$=\frac{3}{2}-e-\frac{1}{2{e}^{2}}$=23​−e−2e21​.