STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

$f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{\log _{e}^{t}}{t+1} d t$

Let $t=\frac{1}{h}, d t=-\frac{1}{h^{2}} d t$

if $t=1, h-1$ or $t=\frac{1}{x}, 4-x$

$f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{\log _{e}\left(\frac{1}{h}\right)}{1+\frac{1}{h}}\left(-\frac{1}{h^{2}} d x\right)$

$f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{-\log h(-1)}{h+1} h d h$

$f\left(\frac{1}{x}\right)=\int_{1}^{x} 9 \frac{\log _{e}^{t}}{t(t+1)} d t   ......(2)$

Adding 1 and 2

$f(x)+f\left(\frac{1}{x}\right)=\int_{1}^{x} \frac{\log _{e}^{t}}{t+1}\left(1+\frac{1}{t}\right) d t$

$\int_{1}^{x} \frac{\log _{e}^{t}}{t+1}\left(1+\frac{1}{t}\right) d t$

$\int_{1}^{x} \frac{\log _{e}^{t}}{t} d t$

$\log t=v$

$\frac{1}{t} d t=d v$

$f(x)+f\left(\frac{1}{x}\right)=\int_{0}^{\log x} v d v$

$=\left[\frac{v^{2}}{2}\right]_{0}^{\log x}$

$=\frac{\left(\log _{e}^{x}\right)^{2}}{2}$

$f(e)+f\left(\frac{1}{e}\right)=\frac{\left(\log _{e}^{e}\right)^{2}}{2}=\frac{1}{2}$