QJEEM21Madef3: JEE Main 2021 Mathematics - Class 12 JEEM21Madef3 Solution: For x > 0, if , then is equal to :

Question

Verified Solution • 2025–26 Session moderate

For x > 0, if

f(x) = \int\limits_1^x {{{{{\log }_e}t} \over {(1 + t)}}dt}

, then

f(e) + f\left( {{1 \over e}} \right)

is equal to :

Accepted Answer

f(x) = \int_1^x {{{\ln t} \over {1 + t}}dt}

then

f\left( {{1 \over x}} \right) = \int_1^{1/x} {{{\ln t} \over {1 + t}}dt}

Let

t = {1 \over u} \Rightarrow dt = - {1 \over {{u^2}}}du

\Rightarrow f\left( {{1 \over x}} \right) = \int_1^x {{{\ln {1 \over u}} \over {1 + {1 \over u}}}\left( { - {1 \over {{u^2}}}} \right)dx}

f\left( {{1 \over x}} \right) = \int_1^x {{{\ln u} \over {u(1 + u)}}} du = \int_1^x {{{\ln t} \over {t(1 + t)}}dt}

\therefore

f(x) + f\left( {{1 \over x}} \right) = \int_1^x {\ln t\left( {{1 \over {1 + t}} + {1 \over {t(1 + t)}}} \right)} dt

= \int_1^x {\ln t\left( {{1 \over {1 + t}} + {1 \over t} - {1 \over {t + 1}}} \right)} dt

= \int_1^x {{{\ln t} \over t}dt = {1 \over 2}{{(\ln x)}^2}}

\therefore

f(e) + f\left( {{1 \over e}} \right) = {1 \over 2}{(\ln e)^2} = {1 \over 2}

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