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$$\int (\arctan x)^2 dx $$

(in terms of elementary functions , if possible)

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up vote 1 down vote accepted

Here is a result by Maple,

$$ 2\,i \left( \arctan \left( x \right) \right) ^{2} \left( {\frac { \left( 1+ix \right) ^{2}}{1+{x}^{2}}}+1 \right) ^{-1}-2\,i \left( \arctan \left( x \right) \right) ^{2}$$ $$ +2\,\arctan \left( x \right) \ln\left( {\frac{ \left( 1+ix \right) ^{2}}{1+{x}^{2}}}+ 1 \right) - i{Li_{2}} \left( -{\frac { \left( 1+ix \right) ^{2}}{1+{x}^{2}} } \right) \,,$$

where the $Li_{s}(z)$ is the polylogarithm function.

You can follow this technique:

Write $\arctan(x)$ in terms of $\ln$ as

$$ \frac{1}{2}\,i \left( \ln \left( 1-ix \right) -\ln \left( 1+ix \right) \right)\,,$$ then use the binomial theorem to expand $ \arctan(x)^2 \,.$

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Can the polylogarithm be avoided ? – Souvik Dey Oct 11 '12 at 8:29
@Souvik:This is not an elementary integral. You have to deal with these kind of functions. – Mhenni Benghorbal Oct 11 '12 at 8:48
@ Mhenni Benghorbal:Thanks for verifying that it's not elementary. – Souvik Dey Oct 11 '12 at 9:05
@Souvik:you are welcome. – Mhenni Benghorbal Oct 11 '12 at 9:06

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