log-x-1-1-log-x-2-2-dx-x-log-x-2-1-C-

Question Number 44604 by arvinddayama01@gmail.com last updated on 02/Oct/18

\int {[\frac{\log x - 1}{1 + {(\log x)}^{2}}]}^{2} dx = \frac{x}{{(\log x)}^{2} + 1} + C

Commented by prof Abdo imad last updated on 02/Oct/18

l e t φ (x) = \frac{x}{1 + {(l n x)}^{2}} w e h a v e

φ^{^{'}} (x) = \frac{1 + {(l n x)}^{2} - x (2 l n (x) \frac{1}{x})}{{(1 + {(l n (x))}^{2})}^{2}}

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

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

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18

\frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}

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

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

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

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

\frac{x}{1 + {(l n x)}^{2}} + c

Commented by arvinddayama01@gmail.com last updated on 02/Oct/18

Thanks