Calculate-lim-x-0-1-x-lnt-1-t-dt-x-1-2-

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

C a l c u l a t e :: \underset{x \to 0^{+}}{l i m} \frac{\int_{1}^{x} \frac{l n t}{1 + t} d t}{{(x - 1)}^{2}} = ?

Answered by puissant last updated on 23/Jun/22

= \int_{1}^{0} \frac{l n t}{1 + t} d t = - \int_{0}^{1} \frac{l n t}{1 + t} d t = \frac{π^{2}}{12}

Answered by Mathspace last updated on 23/Jun/22

= {l i m}_{x \to 1} \frac{\int_{1}^{x} \frac{l n t}{1 + t} d t}{{(x - 1)}^{2}}

= {l i m}_{x \to 1} \frac{l n x}{(1 + x) 2 (x - 1)}

= {l i m}_{x \to 1} \frac{1}{2 (x + 1)} \times \frac{l n x}{x - 1}

= \frac{1}{4} \times 1 = \frac{1}{4}