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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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$\large\frac{d}{dx} \int \limits_{x^2}^{x^3} \bigg(\large\frac {1}{lnt} \bigg)$$dx$

\[\begin {array} {1 1} (a)\;\frac{1}{lnx} (x^2-x) \\ (b)\;\frac{1}{(lnx)}^2. (x^2-x) \\ (c)\;\frac{1}{(lnx)} \{x^3-x^2\} \\ (d)\;none \end {array}\]

Can you answer this question?
 
 

1 Answer

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By using levni theorem
$\large\frac{d}{dx}$$ \int \limits_{g(x)} ^{h(x)} f(t) dt = f(upper\;limit). \large\frac{d}{dx} $$ (upper\;limit)$$-f(lower\;limit). \large\frac{d}{dx}$$ (lower\;limit)$
$\qquad= \large\frac{1}{ln (x^3)} . \frac{d}{dx}(x^3) - \large\frac{1}{(lnx^2)}. \large\frac{d}{dx} (x^2)$
$\qquad= \large\frac{3x^2}{ln(x^3)} -\frac{2x}{lnx^2}$
$\qquad= \large\frac{1}{ln x}. $$[x^2-x]$
Hence a is the correct answer.
answered Dec 14, 2013 by meena.p
 

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