<p>Evaluate \(\displaystyle\int_1^e\frac{dx}{x(1+\ln x)^2}\) [JEE Main 2019]</p>
Step-by-Step Solution
Key Concept: Let t = 1+ln x, dt = dx/x. Limits: x=1\to t=1; x=e\to t=2. \int_1^2 dt/t^2 = [-1/t]_1^2 = -1/2+1 = 1/2.
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<p>Let $t=1+\ln x\Rightarrow dt=dx/x$. Limits: $x=1\to t=1$; $x=e\to t=2$.</p>
<p>$$I=\int_1^2\frac{dt}{t^2}=\left[-\frac{1}{t}\right]_1^2=-\frac{1}{2}+1=\boxed{\frac{1}{2}}$$</p>
Correct Answer: A