For $x>0$, if $f(x)=\int_{1}^{x} \frac{\log _{e} t}{(1+t)} d t$, then $f(e)+f\left(\frac{1}{e}\right)$ is equal to :

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Definite Integrals

For

x > 0

, if

f (x) = \int_{1}^{x} \frac{l o g _{e} t}{( 1 + t )} d t

, then

f (e) + f (\frac{1}{e})

is equal to :

Answer: (B)

f (\frac{1}{e}) = \int_{1}^{1/ e} \frac{lo g _{e} t}{1 + t} d t

substitute

t = \frac{1}{u}

, so

d t = - \frac{d u}{u ^{2}}

and the limits

t = 1

t = \frac{1}{e}

become

u = 1

u = e

f (\frac{1}{e}) = \int_{1}^{e} \frac{- lo g _{e} u}{1 + \frac{1}{u}} (- \frac{d u}{u ^{2}}) = \int_{1}^{e} \frac{lo g _{e} u}{u ( u + 1 )} d u

Splitting

\frac{1}{u ( u + 1 )} = \frac{1}{u} - \frac{1}{u + 1}

separates the second piece as the original integral:

f (\frac{1}{e}) = \int_{1}^{e} \frac{lo g _{e} u}{u} d u - \int_{1}^{e} \frac{lo g _{e} u}{1 + u} d u = [\frac{( lo g _{e} u ) ^{2}}{2}]_{1}^{e} - f (e) = \frac{1}{2} - f (e)

Therefore

f (e) + f (\frac{1}{e}) = \frac{1}{2}

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option