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Question Number 27183 by abdo imad last updated on 02/Jan/18

$$findthevalueofI={\int }_0^1\frac{t-1}{lnt}dt.$$

Answered by prakash jain last updated on 03/Jan/18

$$F\left(x\right)={\int }_0^1\frac{t^x(t-1)}{\ln t}dt$$$$\frac{dF}{dx}={\int }_0^1\frac{\partial }{\partial x}\left(\frac{t^x(t-1)}{\ln t}\right)dt$$$$\frac{dF}{dx}={\int }_0^1\frac{t^x(t-1)}{\ln t}\ln tdt$$$$\frac{dF}{dx}={\int }_0^1{t}^{x+1}-t^{x}dt={[\frac{t^{x+2}}{x+2}-\frac{t^{x+1}}{x+1}]}_0^1$$$$=\frac{1}{x+2}-\frac{1}{x+1}$$$$F\left(x\right)=\ln \mid \frac{x+2}{x+1}\mid +C$$$$F(-\mathrm{\infty })=0\Rightarrow C=0$$$$F\left(x\right)=\ln \mid \frac{x+2}{x+1}\mid$$$$I=F\left(0\right)=\ln 2$$

Commented by prakash jain last updated on 03/Jan/18

$$\mathrm{please}\mathrm{comment}\mathrm{if}\mathrm{mistakes}.\mathrm{thanks}$$

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