If-f-x-1-x-log-t-1-t-dt-then-f-x-f-1-x-1-2-log-x-2-

Question Number 52948 by gunawan last updated on 15/Jan/19

If f (x) = \underset{1}{\int^{x}} \frac{\log t}{1 + t} d t, then

f (x) + f (\frac{1}{x}) = \frac{1}{2} {(\log x)}^{2}

Commented by maxmathsup by imad last updated on 15/Jan/19

l e t φ (x) = f (x) + f (\frac{1}{x}) \Rightarrow φ^{^{'}} (x) = f^{^{'}} (x) - \frac{1}{x^{2}} f^{^{'}} (\frac{1}{x})

= \frac{l o g x}{1 + x} - \frac{1}{x^{2}} \frac{l o g (\frac{1}{x})}{1 + \frac{1}{x}} = \frac{l o g x}{1 + x} + \frac{l o g x}{x^{2} + x} = \frac{x l o g x + l o g x}{x^{2} + x} = \frac{(x + 1) l o g x}{x (x + 1)} = \frac{l o g x}{x}

f r o m a n o t h e r s i d e w e h a v e \frac{d}{d x} (\frac{1}{2} {(l o g x)}^{2}) = \frac{1}{2} \frac{2 l o g x}{x} = \frac{l o g x}{x} \Rightarrow

φ (x) = \frac{1}{2} {(l o g x)}^{2} + c b u t φ (1) = 0 = c \Rightarrow φ (x) = \frac{1}{2} {(l o g x)}^{2} \Rightarrow

f (x) + f (\frac{1}{x}) = \frac{1}{2} {(l o g x)}^{2} .

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

f (x) = \int_{1}^{x} \frac{l n t}{1 + t} d t

f (\frac{1}{x}) = \int_{1}^{\frac{1}{x}} \frac{l n t}{1 + t} d t

k = \frac{1}{t} t = \frac{1}{k} d t = - \frac{1}{k^{2}} d k

\int_{1}^{x} \frac{l n (\frac{1}{k})}{1 + \frac{1}{k}} \times \frac{- d k}{k^{2}}

\int_{1}^{x} \frac{- l n k}{1 + k} \times \frac{k}{- k^{2}} d k

f (\frac{1}{x}) = \int_{1}^{x} \frac{l n k}{1 + k} \times \frac{d k}{k}

f (x) + f (\frac{1}{x}) = \int_{1}^{x} \frac{l n t}{1 + t} + \frac{l n t}{1 + t} \times \frac{d t}{t}

= \int_{1}^{x} \frac{l n t}{1 + t} \times (1 + \frac{1}{t}) d t

= \int_{1}^{x} \frac{l n t}{t}

=∣ \frac{{(l n t)}^{2}}{2} ∣_{1}^{x}

= \frac{{(l n x)}^{2}}{2} a n s w e r

p l s s e e b e l o w \dots

I = \int \frac{l n t}{t} d t

= l n t \times l n t - \int \frac{1}{t} \times l n t d t

2 I = {(l n t)}^{2} s o I = \frac{{(l n t)}^{2}}{2}

Commented by gunawan last updated on 16/Jan/19

thank you very much Sir

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19

m o s t w e l c o m e \dots

Answered by Tinkutara last updated on 15/Jan/19

Commented by Tinkutara last updated on 15/Jan/19