The integral -1e xe 2x- ex x logex dx is equal to-

I=∫_1^e{ ( xe )^2x ( ex )^x }log_ex dx =∫_1^e ( xe )^2xlog_ex dx ∫ nbsp; _1^e ( ex )^x log_ex dx Let ( xe )^2x=t ⇒ 2 ( xe )^2xlog x dx=dt and ( ex )^x=u ...

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Standard XIII

Mathematics

Integration by Substitution

The integral ...

Question

The integral $e \int 1 {{(\frac{x}{e})}^{2 x} - {(\frac{e}{x})}^{x}} {log}_{e} x d x$ is equal to:

\frac{3}{2} - \frac{1}{e} - \frac{1}{2 e^{2}}

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\frac{3}{2} - e - \frac{1}{2 e^{2}}

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- \frac{1}{2} + \frac{1}{e} - \frac{1}{2 e^{2}}

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\frac{1}{2} - e - \frac{1}{e^{2}}

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Solution

The correct option is B $\frac{3}{2} - e - \frac{1}{2 e^{2}}$
$I = e \int 1 {{(\frac{x}{e})}^{2 x} - {(\frac{e}{x})}^{x}} {log}_{e} x d x$

$= e \int 1 {(\frac{x}{e})}^{2 x} {log}_{e} x d x - e \int 1 {(\frac{e}{x})}^{x} {log}_{e} x d x$

Let ${(\frac{x}{e})}^{2 x} = t \Rightarrow 2 {(\frac{x}{e})}^{2 x} log x d x = d t$
and ${(\frac{e}{x})}^{x} = u \Rightarrow - {(\frac{e}{x})}^{x} log x d x = d u$

$∴ I = \frac{1}{2} 1 \int 1 / e^{2} d t + 1 \int e d u$
$= \frac{1}{2} (1 - \frac{1}{e^{2}}) + (1 - e)$
$= \frac{3}{2} - e - \frac{1}{2 e^{2}}$

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The integral e∫1{(xe)2x−(ex)x}logex dx is equal to:

The correct option is B 32−e−12e2I=e∫1{(xe)2x−(ex)x}logex dx =e∫1(xe)2xlogex dx−e∫1(ex)xlogex dx Let (xe)2x=t ⇒ 2(xe)2xlogx dx=dt and (ex)x=u ⇒ −(ex)xlogx dx=du ∴I=121∫1/e2dt+1∫edu =12(1−1e2)+(1−e) =32−e−12e2

The integral $e \int 1 {{(\frac{x}{e})}^{2 x} - {(\frac{e}{x})}^{x}} {log}_{e} x d x$ is equal to: