જો F(x) = f(x) + fleft(frac{1}{x}right),જ્યાં | vedclass

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(C) આપેલ છે $f(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt$. $f\left(\frac{1}{x}\right) = \int_{1}^{1/x} \frac{\log_{e} t}{1+t} dt$ ધ્યાનમાં લો. ધારો

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(C) આપેલ છે $f(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt$. $f\left(\frac{1}{x}\right) = \int_{1}^{1/x} \frac{\log_{e} t}{1+t} dt$ ધ્યાનમાં લો. ધારો કે $t = \frac{1}{u}$,તો $dt = -\frac{1}{u^2} du$. જ્યારે $t=1, u=1$. જ્યારે $t=1/x, u=x$. $f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log_{e}(1/u)}{1+(1/u)} \left(-\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{-\log_{e} u}{\frac{u+1}{u}} \left(\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{-\log_{e} u}{u(u+1)} du$. હવે,$F(x) = f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt + \int_{1}^{x} \frac{-\log_{e} t}{t(1+t)} dt$. $F(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} \left( 1 - \frac{1}{t} \right) dt = \int_{1}^{x} \frac{\log_{e} t}{t} dt$. $F(x) = \left[ \frac{(\log_{e} t)^2}{2} \right]_{1}^{x} = \frac{(\log_{e} x)^2}{2}$. $x=e$ માટે,$F(e) = \frac{(\log_{e} e)^2}{2} = \frac{1}{2} = 0.5$.

જો $F(x) = f(x) + f\left(\frac{1}{x}\right)$,જ્યાં $f(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt$ હોય,તો $F(e) = $

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