Question
$\int_{}^{} {\frac{{dx}}{{\sqrt x \,(x + 9)}}dx} $का मान है
$\sqrt x = t$, रखकर दोनों तरफ वर्ग करने पर $x = {t^2}$
$dx = 2tdt$
$\therefore $$I = 2\int_{}^{} {\frac{{dt}}{{{t^2} + {3^2}}}} = \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{t}{3}} \right)$ ==> $I = \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{{\sqrt x }}{3}} \right)$.
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