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Question Number 172064 by qaz last updated on 23/Jun/22

$$Calculate::\underset{x\to 0^{+}}{lim}\frac{{\int }_1^x\frac{lnt}{1+t}dt}{{(x-1)}^2}=?$$

Answered by puissant last updated on 23/Jun/22

$$={\int }_1^0\frac{lnt}{1+t}dt=-{\int }_0^1\frac{lnt}{1+t}dt=\frac{{\pi }^2}{12}$$

Answered by Mathspace last updated on 23/Jun/22

$$={lim}_{x\to 1}\frac{{\int }_1^x\frac{lnt}{1+t}dt}{{(x-1)}^2}$$$$={lim}_{x\to 1}\frac{lnx}{(1+x)2(x-1)}$$$$={lim}_{x\to 1}\frac{1}{2(x+1)}\times \frac{lnx}{x-1}$$$$=\frac{1}{4}\times 1=\frac{1}{4}$$

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