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Question Number 175077 by rexford last updated on 18/Aug/22

$$Showthat{\int }_0^1\frac{x^b-x^a}{lnx}=ln\left(\frac{b+1}{a+1}\right)$$

Answered by Ar Brandon last updated on 18/Aug/22

$$f\left(b\right)={\int }_0^1\frac{x^b}{\ln x}dx\Rightarrow f{}^{\prime }\left(b\right)={\int }_0^1{x}^{b}dx=\frac{1}{b+1}$$$$\Rightarrow f\left(b\right)=\int \frac{1}{b+1}dx=\ln \mid b+1\mid +C=\ln \mid C_1(b+1)\mid$$$$\Rightarrow f\left(a\right)={\int }_0^1\frac{x^a}{\ln x}dx=\ln \mid C_2(a+1)\mid$$$$\Rightarrow f\left(b\right)-f\left(a\right)={\int }_0^1\frac{x^b-x^a}{\ln x}dx=\ln \mid K\cdot \frac{b+1}{a+1}\mid ,K=\frac{C_1}{C_2}$$$$\Rightarrow f\left(b\right)-f\left(b\right)=0=\ln \mid K\cdot \frac{b+1}{b+1}\mid =\ln K\Rightarrow K=1$$$$\Rightarrow {\int }_0^1\frac{x^b-x^a}{\ln x}dx=\ln \mid \frac{b+1}{a+1}\mid$$

Commented by rexford last updated on 24/Aug/22

$$Thankyou$$

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