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Question Number 175077 by rexford last updated on 18/Aug/22
Show that ∫_0 ^1 ((x^b −x^a )/(lnx))=ln(((b+1)/(a+1)))
$${Show}\:{that}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{b}} −{x}^{{a}} }{{lnx}}={ln}\left(\frac{{b}+\mathrm{1}}{{a}+\mathrm{1}}\right) \\ $$
Answered by Ar Brandon last updated on 18/Aug/22
f(b)=∫_0 ^1 (x^b /(lnx))dx ⇒f ′(b)=∫_0 ^1 x^b dx=(1/(b+1))  ⇒f(b)=∫(1/(b+1))dx=ln∣b+1∣+C=ln∣C_1 (b+1)∣  ⇒f(a)=∫_0 ^1 (x^a /(lnx))dx=ln∣C_2 (a+1)∣  ⇒f(b)−f(a)=∫_0 ^1 ((x^b −x^a )/(lnx))dx=ln∣K∙((b+1)/(a+1))∣, K=(C_1 /C_2 )  ⇒f(b)−f(b)=0=ln∣K∙((b+1)/(b+1))∣=lnK⇒K=1  ⇒∫_0 ^1 ((x^b −x^a )/(lnx))dx=ln∣((b+1)/(a+1))∣
$${f}\left({b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{b}} }{\mathrm{ln}{x}}{dx}\:\Rightarrow{f}\:'\left({b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{b}} {dx}=\frac{\mathrm{1}}{{b}+\mathrm{1}} \\ $$$$\Rightarrow{f}\left({b}\right)=\int\frac{\mathrm{1}}{{b}+\mathrm{1}}{dx}=\mathrm{ln}\mid{b}+\mathrm{1}\mid+{C}=\mathrm{ln}\mid{C}_{\mathrm{1}} \left({b}+\mathrm{1}\right)\mid \\ $$$$\Rightarrow{f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}} }{\mathrm{ln}{x}}{dx}=\mathrm{ln}\mid{C}_{\mathrm{2}} \left({a}+\mathrm{1}\right)\mid \\ $$$$\Rightarrow{f}\left({b}\right)−{f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{b}} −{x}^{{a}} }{\mathrm{ln}{x}}{dx}=\mathrm{ln}\mid{K}\centerdot\frac{{b}+\mathrm{1}}{{a}+\mathrm{1}}\mid,\:{K}=\frac{{C}_{\mathrm{1}} }{{C}_{\mathrm{2}} } \\ $$$$\Rightarrow{f}\left({b}\right)−{f}\left({b}\right)=\mathrm{0}=\mathrm{ln}\mid{K}\centerdot\frac{{b}+\mathrm{1}}{{b}+\mathrm{1}}\mid=\mathrm{ln}{K}\Rightarrow{K}=\mathrm{1} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{b}} −{x}^{{a}} }{\mathrm{ln}{x}}{dx}=\mathrm{ln}\mid\frac{{b}+\mathrm{1}}{{a}+\mathrm{1}}\mid \\ $$
Commented by rexford last updated on 24/Aug/22
Thank you
$${Thank}\:{you} \\ $$

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