The integral ∫xex+exxlogex dx is equal to

∫xex+exxlogexdx=xex=t xexx1x+logx-1dx=dt xex1+logx-1dx=dt xexlogxdx=dtexx=s exxx-1x+logexdx=ds exx-1+1-logxdx=ds -exxlogxdx=ds ∫xex+exxlogexdx =∫dt-∫ds =t-s+C xex-exx+C*Note: Given question and options are wrong *∫x2x+2xxlog2 xdxx2x−2xx+C

The integral ∫xex+exxlogex dx is equal to

∫xex+exxlogexdx=xex=t xexx1x+logx-1dx=dt xex1+logx-1dx=dt xexlogxdx=dtexx=s exxx-1x+logexdx=ds exx-1+1-logxdx=ds -exxlogxdx=ds ∫xex+exxlogexdx =∫dt-∫ds =t-s+C xex-exx+C*Note: Given question and options are wrong *∫x2x+2xxlog2 xdxx2x−2xx+C

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The integral $\int ({(\frac{x}{e})}^{x} + {(\frac{e}{x})}^{x}) \log_{e} x d x$ is equal to

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${(\frac{x}{e})}^{x} \log_{2} (\frac{x}{e}) + C$

${(\frac{x}{e})}^{x} \log_{2} (\frac{e}{x}) + C$

${(\frac{x}{e})}^{x} - {(\frac{e}{x})}^{x} + C$

${(\frac{x}{e})}^{x} + {(\frac{e}{x})}^{x} + C$

answer is C.

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Detailed Solution

$\int [{(\frac{x}{e})}^{x} + {(\frac{e}{x})}^{x}] \log_{e} x dx =$
${(\frac{x}{e})}^{x} = t {(\frac{x}{e})}^{x} (x \frac{1}{x} + \log x - 1) d x = d t {(\frac{x}{e})}^{x} (1 + \log x - 1) d x = d t {(\frac{x}{e})}^{x} \log x d x = d t$
${(\frac{e}{x})}^{x} = s {(\frac{e}{x})}^{x} (x (- \frac{1}{x}) + \log (\frac{e}{x})) d x = d s {(\frac{e}{x})}^{x} (- 1 + 1 - \log x) d x = d s - {(\frac{e}{x})}^{x} \log x d x = d s$

$\int [{(\frac{x}{e})}^{x} + {(\frac{e}{x})}^{x}] \log_{e}^{x} d x = \int d t - \int d s = t - s + C {(\frac{x}{e})}^{x} - {(\frac{e}{x})}^{x} + C$
*Note: Given question and options are wrong *
$\int ({(\frac{x}{2})}^{x} + {(\frac{2}{x})}^{x}) \log_{2}^{} x d x$
${(\frac{x}{2})}^{x} - {(\frac{2}{x})}^{x} + C$