Question-156933

Question Number 156933 by CAIMAN last updated on 17/Oct/21

Answered by mindispower last updated on 17/Oct/21

\frac{1}{t} = x

\int_{\frac{1}{2}}^{\frac{1}{3}} \frac{l n (x)}{x (1 + x)} d x = \int \frac{l n (x)}{x} - \frac{l n (x)}{1 + x} d x

= \frac{{l n}^{2} (x)}{2} - l n (x) l n (1 + x) + \int \frac{l n (1 + x)}{x} d x

y = - x i n \int \frac{l n (1 + x)}{x} d x

= \int \frac{l n (1 - y)}{y} d y = - {L i}_{2} (y), {L i}_{2} \dots d i l o g a r i t h m f u n c t i o n

\int \frac{l n (x)}{x (1 + x)} = \frac{{l n}^{2} (x)}{2} - l n (x) l n (1 + x) - {L i}_{2} (- x) + c

\int_{\frac{1}{2}}^{\frac{1}{3}} \frac{l n (x)}{x (1 + x)} d x = \frac{1}{2} ({l n}^{2} (3) - {l n}^{2} (3)) - l n (\frac{1}{3}) l n (\frac{4}{3}) + l n (\frac{1}{2}) l n (\frac{3}{2})

- {l i}_{2} (- \frac{1}{3}) + {l i}_{2} (- \frac{1}{2})