Question:
The integral
\(\int_1^2 e^x \cdot x^x\,(2+\log_e x)\,dx\)
equals :
The integral
\(\int_1^2 e^x \cdot x^x\,(2+\log_e x)\,dx\)
equals :
\(\int_1^2 e^x \cdot x^x\,(2+\log_e x)\,dx\)
equals :
Show Hint
Look for a function whose derivative is \(e^x x^x(2+\log_e x)\); try \(e^x x^x\).
Updated On: Nov 5, 2026
- \(e(4e+1)\)
- \(e(2e-1)\)
- \(e(4e-1)\)
- \(4e^2-1\)
Show Solution
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The Correct Option is C
Solution and Explanation
Let \(g(x)=e^x x^x\). Then \(\ln g = x+x\ln x\).
Differentiate: \(g'(x)/g(x)=1+(\ln x+1)=2+\ln x\).
So \(g'(x)=e^x x^x(2+\ln x)\), i.e. the integrand is exactly \(g'(x)\).
\(I=[e^x x^x]_1^2 = e^2\cdot 2^2 - e\cdot 1^1 = 4e^2-e = e(4e-1)\).
Answer: (3) \(e(4e-1)\)
Differentiate: \(g'(x)/g(x)=1+(\ln x+1)=2+\ln x\).
So \(g'(x)=e^x x^x(2+\ln x)\), i.e. the integrand is exactly \(g'(x)\).
\(I=[e^x x^x]_1^2 = e^2\cdot 2^2 - e\cdot 1^1 = 4e^2-e = e(4e-1)\).
Answer: (3) \(e(4e-1)\)
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