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Question Number 45117 by rahul 19 last updated on 09/Oct/18

$$Provethat{\int }_0^1\frac{x^a-1}{\log x}dx=\log (a+1).$$

Commented by maxmathsup by imad last updated on 09/Oct/18

$$fora>-1letf\left(a\right)={\int }_0^1\frac{x^a-1}{ln\left(x\right)}dxchangementln\left(x\right)=-tgive$$$$f\left(a\right)=-{\int }_0^{+\mathrm{\infty }}\frac{e^{-at}-1}{-t}(-1)e^{-t}dt=-{\int }_0^{\mathrm{\infty }}\frac{e^{-(a+1)t}-e^{-t}}{t}dt$$$$={\int }_0^{\mathrm{\infty }}\frac{e^{-t}-e^{-(a+1)t}}{t}dt\Rightarrow f^{{}^{\prime }}\left(a\right)={\int }_0^{\mathrm{\infty }}{e}^{-(a+1)t}dt={[-\frac{1}{a+1}{e}^{-(a+1)t}]}_{t=0}^{+\mathrm{\infty }}$$$$=\frac{1}{a+1}\Rightarrow f\left(a\right)=ln\mid a+1\mid +c=ln(a+1)+cbut$$$$f\left(0\right)=0=ln\left(1\right)+c\Rightarrow c=0\Rightarrow f\left(a\right)=ln(a+1).$$$$$$$$$$

Commented by rahul 19 last updated on 09/Oct/18

thanks prof Abdo ☺️

Commented by turbo msup by abdo last updated on 09/Oct/18

$$youarewelcomesir.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18

$$I\left(a\right)={\int }_0^1\frac{x^a-1}{lnx}dx$$$$\frac{dI}{da}={\int }_0^1\frac{1}{lnx}\frac{\partial }{\partial a}(x^a-1)dx$$$$={\int }_0^1\frac{1}{lnx}\times x^a\times lnxdx$$$$={\int }_0^1{x}^{a}dx$$$$=\mid \frac{x^{a+1}}{a+1}{\mid }_0^1$$$$=\frac{1}{a+1}$$$$\int dI=\int \frac{da}{a+1}I=ln(a+1)$$

Commented by rahul 19 last updated on 09/Oct/18

Sir , how will this come to my mind first we have to differentiate!! ����

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18

$$ishallpostsomeexample...thistypeintregal$$$$areadvancedintregal...frombookishallpost..$$

Commented by rahul 19 last updated on 12/Oct/18

$$Sir,plsspost2-3examples.....$$

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