Question
$\int_{}^{} {\frac{{dx}}{{2\sqrt x (1 + x)}} = } $
$\sqrt x \, = t$==> $\frac{1}{{2\sqrt x }}dx = dt$ रखने पर
$\therefore I = \int {\frac{{dt}}{{1 + {t^2}}}} = {\tan ^{ - 1}}t + c$$ = {\tan ^{ - 1}}(\sqrt x ) + c$.
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$(S2)$ : $13(13)^n-11 \mathrm{n}-13$ अनंत $\mathrm{n} \in \mathrm{N}$ के लिए
$144$ से विभाज्य हैमें से
$\tan ^{-1}(\sqrt{\frac{1-\cos x}{1+\cos x}}), x<\pi$