<p>Evaluate \(\displaystyle\int_0^\infty\frac{\ln x}{1+x^2}\,dx\) [JEE Advanced 2004]</p>
Step-by-Step Solution
Key Concept: Let x=1/t in the half \int_1^\infty: \int_1^\infty lnx/(1+x^2)dx = -\int_0^1 lnx/(1+x^2)dx. So total = 0.
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<p>Split: $I=\int_0^1\frac{\ln x}{1+x^2}dx+\int_1^\infty\frac{\ln x}{1+x^2}dx$.</p>
<p>In second piece, let $x=1/t$: $\int_1^\infty\frac{\ln x}{1+x^2}dx=\int_0^1\frac{-\ln t}{1+1/t^2}\cdot\frac{-dt}{t^2}=\int_0^1\frac{-\ln t}{t^2+1}dt$.</p>
<p>So: $I=\int_0^1\frac{\ln x}{1+x^2}dx-\int_0^1\frac{\ln x}{1+x^2}dx=\boxed{0}$.</p>
Correct Answer: A