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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Find value of $\int \limits_0^1 \large\frac{x^{\alpha}-1}{\log x}$$dx \qquad \alpha \geq 0$

\[\begin {array} {1 1} (a)\;\log _e 2 \\ (b)\;\log _e( 2 + \alpha) \\ (c)\;\log_e(\alpha -1) \\ (d)\;\log_e(1+\alpha) \end {array}\]

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1 Answer

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$f(\alpha) =\int \limits_0^1 \large\frac{x^{\alpha}-1}{\log x } $$dx$------(i)
$\large\frac{dF( \alpha)}{d \alpha} =\int \limits_0^1 \large\frac{d}{d \alpha}- \bigg( \large\frac{x^{\alpha}-1}{\log n}\bigg)$$dx$
$\large\frac{dF( \alpha)}{d \alpha} =\int \limits_0^1 \large\frac{x^{\alpha}. \log x}{\log x}$$dx$
$\qquad = \bigg[\large\frac{x^{\alpha +1}}{\alpha +1} \bigg]_0^1$
$\qquad= \large\frac{1}{\alpha +1}$
$dF(\alpha)=\large\frac{d \alpha}{\alpha+1}$
=> $F( \alpha )= ln (\alpha +1)+c$------(ii)
From (i) $F(0) =\int \limits_0^1 \large\frac{1-1}{\log n }$$dx=0$
From (ii) $F(0) =\log (1) +c=0$
$c=0$
So $F( \alpha)= ln (\alpha +1)$
Hence d is the correct answer.
answered Dec 20, 2013 by meena.p
edited Oct 15, 2014 by sharmaaparna1
 

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