<p>Evaluate \(\displaystyle\int_0^1\tan^{-1}x\,dx\) [JEE Main 2017]</p>
Step-by-Step Solution
Key Concept: IBP: u=tan⁻^1x, dv=dx \to \int_0^1 tan⁻^1x dx = [x \cdot tan⁻^1x]_0^1 - \int_0^1 x/(1+x^2)dx = \pi/4 - (1/2)ln2.
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<p>IBP: $u=\arctan x$, $dv=dx$:</p>
<p>$$\int_0^1\arctan x\,dx=[x\arctan x]_0^1-\int_0^1\frac{x}{1+x^2}dx=\frac{\pi}{4}-\frac{1}{2}[\ln(1+x^2)]_0^1=\frac{\pi}{4}-\frac{\ln 2}{2}$$</p>
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Correct Answer: A