find-the-value-of-I-0-1-t-1-lnt-dt-

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

f i n d t h e v a l u e o f I = \int_{0}^{1} \frac{t - 1}{l n t} d t .

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

F (x) = \int_{0}^{1} \frac{t^{x} (t - 1)}{\ln t} d t

\frac{d F}{d x} = \int_{0}^{1} \frac{\partial}{\partial x} (\frac{t^{x} (t - 1)}{\ln t}) d t

\frac{d F}{d x} = \int_{0}^{1} \frac{t^{x} (t - 1)}{\ln t} \ln t d t

\frac{d F}{d x} = \int_{0}^{1} t^{x + 1} - t^{x} d t = {[\frac{t^{x + 2}}{x + 2} - \frac{t^{x + 1}}{x + 1}]}_{0}^{1}

= \frac{1}{x + 2} - \frac{1}{x + 1}

F (x) = \ln ∣ \frac{x + 2}{x + 1} ∣ + C

F (- \infty) = 0 \Rightarrow C = 0

F (x) = \ln ∣ \frac{x + 2}{x + 1} ∣

I = F (0) = \ln 2

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

please comment if mistakes . thanks