求不定积分

\[ \int{(\frac{\arctan x}{x-\arctan x})^2 dx}\]

提示： \( -\frac{v’}{v^2}dv = d(\frac{1}{v})\)

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\[ \begin{align} ~& \int{(\frac{\arctan x}{x-\arctan x})^2 dx} \\
&= \int{\frac{t^2}{(\tan t – t)^2} d(\tan(t))} \\
&= \int{\frac{t^2}{(\cos t( \tan t – t))^2} d(t)} \\
&= \int{\frac{t^2}{( \sin t – t \cos t)^2} d(t)} \\
&= \int{\frac{t^2}{( \sin t – t \cos t)^2} d(t)} \\
&= \int{\frac{t}{\sin t}\frac{ t \sin t}{( \sin t – t \cos t)^2} d(t)} \\
&= \int{\frac{t}{\sin t}d(\frac{1}{\sin t – t \cos t})} \\
&= \frac{t}{\sin t} \frac{1}{\sin t – t \cos t} – \int{\frac{1}{\sin t – t \cos t} d(\frac{t}{\sin t})}\\
&= \frac{t}{\sin t} \frac{1}{\sin t – t \cos t} – \frac{\cos t}{\sin t} + C\\
&= \frac{x \arctan x}{x – \arctan x} + C\\
\end{align} \]

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