Answer
\[ = x{\tan ^{ - 1}}x - \frac{1}{2}\ln \,\left( {1 + {x^2}} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{{\tan }^{ - 1}}xdx} \hfill \\
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set\,\,\,the\,\,substitution \hfill \\
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v = x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,dv = dx \hfill \\
u = {\tan ^{ - 1}}x\,\, \to \,\,\,\,\,du = \frac{1}{{1 + {x^2}}} \hfill \\
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use\,\,uv - \int_{}^{} {vdu} \hfill \\
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{\text{replacing the values }}{\text{in the equation}} \hfill \\
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= x{\tan ^{ - 1}}x - \int_{}^{} {\frac{x}{{1 + {x^2}}}} dx \hfill \\
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rewrite \hfill \\
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x{\tan ^{ - 1}}x - \frac{1}{2}\int_{}^{} {\frac{{2x}}{{1 + {x^2}}}dx} \hfill \\
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integrate\,\,{\text{using }}\int {\frac{{du}}{u} = \ln u + C} \hfill \\
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= x{\tan ^{ - 1}}x - \frac{1}{2}\ln \,\left( {1 + {x^2}} \right) + C \hfill \\
\hfill \\
\end{gathered} \]
