Option (B)

Explanation:

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\int \mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x}) \mathrm{dx}=\mathrm{f}(\mathrm{x}) \int \mathrm{g}(\mathrm{x}) \mathrm{dx}=\int\left(\mathrm{f}(\mathrm{x}) \int \mathrm{g}(\mathrm{x}) \mathrm{dx}\right) \mathrm{dx} \\

\int_{0}^{1} \mathrm{x} \tan ^{-1} \mathrm{x} \mathrm{dx}=\int_{0}^{1}\left(\tan ^{-1} \mathrm{x}\right) \mathrm{x} \mathrm{dx} \\

=\tan ^{-1} \mathrm{x} \int_{0}^{1} \mathrm{x} \mathrm{dx}-\int_{0}^{1}\left(\frac{\mathrm{d}\left(\tan ^{-1} \mathrm{x}\right)}{\mathrm{dx}} \int \mathrm{x} \cdot \mathrm{dx}\right) \mathrm{dx} \\

=\left[\tan ^{-1} \mathrm{x} \cdot \frac{\mathrm{x}}{2}\right]_{0}^{1}-\int_{0}^{1} \frac{1}{1+\mathrm{x}^{2}} \mathrm{dx} \\

=\left[\frac{\mathrm{x}^{2}}{2} \tan ^{-1} \mathrm{x}\right]-\frac{1}{2} \int_{0}^{1} \frac{\mathrm{x}^{2}+1-1}{\mathrm{x}^{2}+1} \mathrm{dx} \\

=\left[\frac{\mathrm{x}^{2}}{2} \tan ^{-1} \mathrm{x}\right]_{0}^{1}-\frac{1}{2}\left[\int_{0}^{1} \frac{\mathrm{x}^{2}+1}{\mathrm{x}^{2}+1} \mathrm{dx}-\int_{0}^{1} \frac{\mathrm{dx}}{\mathrm{x}^{2}+1}\right] \\

=\left[\frac{\mathrm{x}^{2}}{2} \tan ^{-1} \mathrm{x}\right]-\frac{1}{2}\left[\int_{0}^{1} \mathrm{dx}-\int_{0}^{1} \frac{\mathrm{dx}}{\mathrm{x}^{2}+1}\right] \\

{\left[\mathrm{U} \sin g \int \frac{\mathrm{dx}}{\mathrm{a}^{2}+\mathrm{x}^{2}}=\frac{1}{\mathrm{a}} \tan ^{-1} \frac{\mathrm{x}}{\mathrm{a}}+\mathrm{C}\right]} \\

=\left[\frac{\mathrm{x}^{2}}{2} \tan ^{-1} \mathrm{x}-\frac{1}{2} \mathrm{x}+\frac{1}{2}+\frac{1}{1} \tan ^{-1} \frac{\mathrm{x}}{1}\right]_{0}^{1} \mathrm{C}

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=\left[\frac{x^{2}}{2} \tan ^{-1} x-\frac{x}{2}+\frac{1}{2} \tan ^{-1} x\right]_{0}^{1} \\

=\left[\frac{(1)^{2}}{2} \tan ^{1}(1)-\frac{1}{2}+\frac{1}{2} \tan ^{-1} 1\right]-\left[\frac{0^{2}}{2} \tan ^{-1} 0-\frac{0}{2}+\frac{1}{2} \tan ^{-1}(0)\right] \\

=\left[\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}\right]-[0-0+0] \\

=\pi / 4-\frac{1}{2}

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