∫_0 ^1 ((x^(98) −99x+98)/(logx))dx


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Question Number 103871 by Dwaipayan Shikari last updated on 17/Jul/20

$\int_{0}^{1} \frac{x^{98} - 99 x + 98}{l o g x} d x$
Commented by Dwaipayan Shikari last updated on 18/Jul/20

$I (a) = \int_{0}^{1} \frac{x^{a} - 1}{l o g x}$ $I^{^{'}} (a) = \int_{0}^{1} \frac{\partial}{\partial a} (\frac{x^{a} - 1}{l o g x}) = \int_{0}^{1} \frac{x^{a} l o g x}{l o g x} = \int_{0}^{1} x^{a} = {[\frac{x^{a + 1}}{a + 1}]}_{0}^{1} = \frac{1}{a + 1}$ $\int I^{^{'}} (a) = \int \frac{1}{a + 1}$ $I (a) = l o g (a + 1) + C = \int \frac{x^{a} - 1}{l o g x}$ $S o,$ $\int_{0}^{1} \frac{x^{98} - 1}{l o g x} - 99 \int_{0}^{1} \frac{x - 1}{l o g x} = l o g 99 - 99 l o g 2$
Commented by Dwaipayan Shikari last updated on 18/Jul/20

$K i n d l y c h e c k m y a n s w e r$
	Answered by Ar Brandon last updated on 17/Jul/20
	$Let x=e^u ⇒dx=e^u du ⇒I=∫_∞ ^0 ((e^(99u) −99e^(2u) +98e^u )/u)du =−∫_0 ^∞ {Σ_(k=0) ^∞ ((99^k u^(k−1) )/(k!))−99Σ_(k=0) ^∞ ((2^k u^(k−1) )/(k!))+98Σ_(k=0) ^∞ (u^(k−1) /(k!))}du =−[Σ_(k=0) ^∞ ((99^k u^k )/(k(k!)))−99Σ_(k=0) ^∞ ((2^k u^k )/(k(k!)))+98Σ_(k=0) ^∞ (u^k /(k(k!)))]_0 ^∞$
	$Let x = e^{u} \Rightarrow dx = e^{u} du$ $\Rightarrow I = \int_{\infty}^{0} \frac{e^{99 u} - {99 e}^{2 u} + {98 e}^{u}}{u} du$ $= - \int_{0}^{\infty} {\underset{k = 0}{\sum^{\infty}} \frac{99^{k} u^{k - 1}}{k!} - 99 \underset{k = 0}{\sum^{\infty}} \frac{2^{k} u^{k - 1}}{k!} + 98 \underset{k = 0}{\sum^{\infty}} \frac{u^{k - 1}}{k!}} du$ $= - {[\underset{k = 0}{\sum^{\infty}} \frac{99^{k} u^{k}}{k (k!)} - 99 \underset{k = 0}{\sum^{\infty}} \frac{2^{k} u^{k}}{k (k!)} + 98 \underset{k = 0}{\sum^{\infty}} \frac{u^{k}}{k (k!)}]}_{0}^{\infty}$
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