Question:
The integral \(\int_1^e \left\{\left(\frac{x}{e}\right)^{2x}-\left(\frac{e}{x}\right)^{x}\right\}\log_e x\,dx\) is equal to :
The integral \(\int_1^e \left\{\left(\frac{x}{e}\right)^{2x}-\left(\frac{e}{x}\right)^{x}\right\}\log_e x\,dx\) is equal to :
Show Hint
Put \(f(x)=\left(\frac{x}{e}\right)^x\) so that \(f'(x)=f(x)\log_e x\); the integrand becomes \(\left(f^2-\frac{1}{f}\right)\frac{f'}{f}\).
Updated On: Nov 5, 2026
- \(-\frac{1}{2}+\frac{1}{e}-\frac{1}{2e^2}\)
- \(\frac{3}{2}-e-\frac{1}{2e^2}\)
- \(\frac{1}{2}-e-\frac{1}{e^2}\)
- \(\frac{3}{2}-\frac{1}{e}-\frac{1}{2x^2}\)
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The Correct Option is B
Solution and Explanation
Let \(f(x)=\left(\frac{x}{e}\right)^{x}=e^{x\ln x-x}\). Then \(f'(x)=f(x)\cdot\ln x\) (since \(\frac{d}{dx}(x\ln x-x)=\ln x\)).
The integrand is \(\left(f^2-\frac{1}{f}\right)\ln x = f\,f' - \frac{f'}{f^2}\).
So the integral \(=\left[\frac{f^2}{2}+\frac{1}{f}\right]_1^e\).
At \(x=e\): \(f=1\), value \(=\frac12+1=\frac32\). At \(x=1\): \(f=\frac1e\), value \(=\frac{1}{2e^2}+e\).
Result \(=\frac32-e-\frac{1}{2e^2}\).
Answer: (2) \(\frac{3}{2}-e-\frac{1}{2e^2}\)
The integrand is \(\left(f^2-\frac{1}{f}\right)\ln x = f\,f' - \frac{f'}{f^2}\).
So the integral \(=\left[\frac{f^2}{2}+\frac{1}{f}\right]_1^e\).
At \(x=e\): \(f=1\), value \(=\frac12+1=\frac32\). At \(x=1\): \(f=\frac1e\), value \(=\frac{1}{2e^2}+e\).
Result \(=\frac32-e-\frac{1}{2e^2}\).
Answer: (2) \(\frac{3}{2}-e-\frac{1}{2e^2}\)
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