Let $F(x) = f(x) + f\left(\frac{1}{x}\right)$, where $f(x) = \int_{1}^{x} \frac{\log t}{1+t} dt$. Then $F(e)$ equals

JEE Main · 2007 · 29 Apr 2007 · Offline · Definite Integrals

Let

F (x) = f (x) + f (\frac{1}{x})

, where

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

. Then

F (e)

equals

Answer: (A)

Here

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

with

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

. Substituting

t = \frac{1}{u}

in the second integral recasts it as

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

, so that

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

. The bracket simplifies to

\frac{1}{t}

, leaving

F (e) = \int_{1}^{e} \frac{l o g t}{t} d t = [\frac{( l o g t ) ^{2}}{2}]_{1}^{e} = \frac{1}{2}

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option

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