The integral ${\int }_{0}^{1}{\cot }^{-1}\left(1+x+{x}^{2}\right)dx$ is equal to: [JEE Main 2026, 4 Apr (Shift 2)]

Question

YESPYQ · Accepted Answer

Correct answer: d) $2{\tan }^{-1}2-\frac{1}{2}{\log }_{\mathrm{e}}\left(\frac{5}{4}\right)-\frac{\pi }{2}$. $ I=\int_0^1 \cot ^{-1}\left(1+x+x^2\right) d x$ $=\int_0^1 \tan ^{-1}\left(\frac{1}{1+x+x^2}\right) d x$ $=\int_0^1 \tan ^{-1}\left(\frac{x+1-x}{1+x(x+1)}\right) d x$ $I=\int_0^1 \tan ^{-1}(x+1) d x-\int_0^1 \tan ^{-1} x d x$ $=\int_1^2 \tan ^{-1} x d x-\int_0^1 \tan ^{-1} x d x$ $\because \int \tan ^{-1} x d x=x \tan ^{-1} x-\frac{1}{2} \ln \left(1+x^2\right)$ $\Rightarrow I=2 \tan ^{-1} 2-\frac{1}{2} \ln \left(\frac{5}{4}\right)-\frac{\pi}{2}$

The integral \({\int }_{0}^{1}{\cot }^{-1}\left(1+x+{x}^{2}\right)dx\) is equal to: [JEE Main 2026, 4 Apr (Shift 2)]

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