int_0^1 {frac{{{x^b} - 1}}{{log x}}} ,dx&nbsp | vedclass

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Question

(d) Let $I(b) = \int_0^1 {\frac{{{x^b} - 1}}{{\log x}}} dx $

$\Rightarrow I'(b) = \int_0^1 {\frac{{{x^b}\log x}}{{\log x}}dx} $

(If $I(\alph

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એકપણ નહીં.

Vedclass · Answer

(d) Let $I(b) = \int_0^1 {\frac{{{x^b} - 1}}{{\log x}}} dx $

$\Rightarrow I'(b) = \int_0^1 {\frac{{{x^b}\log x}}{{\log x}}dx} $

(If $I(\alpha ) = \int_0^b {f(x,\alpha )dx} $, then $I'(\alpha ) = \int_0^b {f'(x,\alpha )dx} $,

where $f'(x,\alpha )$ is derivative of $f(x,\alpha )$ w.r.t. $\alpha $ keeping $x$ constant)

$I'(b) = \int_0^1 {{x^b}dx = \frac{1}{{b + 1}}} $

==> $I(b) = \int {\frac{{db}}{{b + 1}} + c = \log (b + 1) + c} $

If $b = 0$, then $I(b) = 0$,

so $c = 0$==>$I(b) = \log (b + 1)$.

$\int_0^1 {\frac{{{x^b} - 1}}{{\log x}}} \,dx = . . . ..$

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