calculste lim_(x→0) ∫_x ^(2x) ((arctan(xt))/(t+x))dt


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Question Number 93908 by abdomathmax last updated on 16/May/20

$c a l c u l s t e {l i m}_{x \to 0} \int_{x}^{2 x} \frac{a r c t a n (x t)}{t + x} d t$
Commented by redmiiuser last updated on 16/May/20

$A n s w e r : 0$
Commented by abdomathmax last updated on 17/May/20

$l e t F (x) = \int_{x}^{2 x} \frac{a r c t a n (x t)}{t + x} d t w e d o t h e c h a n g e m e n t$ $x t = u \Rightarrow F (x) = \int_{x^{2}}^{2 x^{2}} \frac{a r c t a n (u)}{\frac{u}{x} + x} \frac{d u}{x}$ $= \int_{x^{2}}^{2 x^{2}} \frac{a r c t a n u}{u + x^{2}} d u \exists c_{x} \in] x^{2}, 2 x^{2} [/$ $F (x) = a r c t s n (c_{x}) \int_{x^{2}}^{2 x^{2}} \frac{d u}{u + x^{2}} = a r c t a n (c_{x}) {[l n (u + x^{2})]}_{x^{2}}^{2 x^{2}}$ $= a r c t a n (c_{x}) \times (l n (\frac{3 x^{2}}{2 x^{2}})) b u t x \to 0 \Rightarrow c_{x} \to 0 \Rightarrow$ ${l i m}_{x \to 0} F (x) = 0 \times l n (\frac{3}{2}) = 0$
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