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### $\int\limits_0^1 {\frac{{{x^{e - 1}} - 1}}{{\ln x}}dx} = ?$

Using the result,

For $I(\lambda ) = \int\limits_a^b {f(x,\lambda )dx}$, $\frac{{dI(\lambda )}}{{d\lambda }} = \int\limits_a^b {\left[ {\frac{\partial }{{\partial \lambda }}f(x,\lambda )} \right]dx}$

Let, $I(\lambda ) = \int\limits_0^1 {\frac{{{x^\lambda } - 1}}{{\ln x}}dx}$

$\frac{{dI(\lambda )}}{{d\lambda }} = \int\limits_0^1 {\left[ {\frac{\partial }{{\partial \lambda }}\left( {\frac{{{x^\lambda } - 1}}{{\ln x}}} \right)} \right]dx}$

$\Rightarrow \frac{{dI(\lambda )}}{{d\lambda }} = \int\limits_0^1 {\frac{{{x^\lambda }\ln x}}{{\ln x}}dx}$

$= \left. {\frac{{{x^{\lambda + 1}}}}{{\lambda + 1}}} \right|_0^1 = \frac{1}{{\lambda + 1}}$

$I(\lambda ) = \int {\frac{1}{{\lambda + 1}}} d\lambda = \ln (\lambda + 1) + C$

$I(0) = C$

Also, $I(0) = \int\limits_0^1 {\frac{{{x^0} - 1}}{{\ln x}}dx} = 0$

So, $C = 0$

$\therefore I(\lambda ) = \ln (\lambda + 1)$

The given integral $= I(e - 1) = \ln (e - 1 + 1) = 1$

### 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 = ${}^{12}{C_6} = 924$