<p>Evaluate \(\displaystyle\int_1^e(\ln x)^2\,dx\) [JEE Main 2019]</p>
Step-by-Step Solution
Key Concept: IBP twice: let u=(ln x)^2, dv=dx. Result: [x(ln x)^2]_1^e - 2\int_1^e ln x dx = e - 2(e-1) ... = e-2.
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<p>$\int_1^e(\ln x)^2dx=[x(\ln x)^2]_1^e-2\int_1^e\ln x\,dx$</p>
<p>$=e\cdot1-0-2[x\ln x-x]_1^e=e-2[(e-e)-(0-1)]=e-2\cdot1=e-2$</p>
Correct Answer: A