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### Evaluate, $\int\limits_{ - 2/3}^0 {\frac{{dx}}{{{{(x + 1)}^{1/2}} + {{(x + 1)}^{3/2}}}}}$

Let, $\sqrt {x + 1} = t$

Hence, $\frac{1}{{2\sqrt {x + 1} }}dx = dt$

Thus, $dx = 2\sqrt {x + 1} dt$

The given integral = $\int\limits_{1/\sqrt 3 }^1 {\frac{{2\sqrt {x + 1} dt}}{{{{(x + 1)}^{1/2}} + {{(x + 1)}^{3/2}}}}}$

=$\int\limits_{1/\sqrt 3 }^1 {\frac{{2tdt}}{{t + {t^3}}}}$

=$2\int\limits_{1/\sqrt 3 }^1 {\frac{{dt}}{{1 + {t^2}}}}$

=$2\left[ {{{\tan }^{ - 1}}t|_{1/\sqrt 3 }^1} \right]$

=$2\left( {\frac{\pi }{4} - \frac{\pi }{6}} \right) = \frac{\pi }{6}$

### Sum of the coefficients in the expansion of $(x+y)^n$ ....

If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$