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Integrale di arctan(2x) dx ?

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∫arctan(2x) dx

Poniamo:

g'(x)=1 e f(x)= arctan(2x)

∫arctan(2x) dx =x (arctang(2x))-∫x [x/(1+4x^2)] (2) dx =x (arctang(2x))-1/4 ∫8x/(4x^2+1) dx=

=x (arctang(2x))-1/4 ln(4x^2+1) + c

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§arctg(2x)= xarctg(2x) - § x(2/(1+4x^2)= // - 1/4 § 8x/(4x^2 +1)= // - log(4x^2+1)=

= xarctg(2x) - log(4x^2+1)/4 +c

Bravo Bruno :D

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