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Definite Integrals Part 2

Section: Mathematics

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Question : 97 of 100

Marks: +1, -0

For

x > 0

f(x)=\int\limits_{1}^{x} \frac{\log_e t}{1+t} \, dt

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

[26 Feb 2021 Shift 2]

1
$-1$
$\frac{1}{2}$
0

Solution:

f(x)=\int\limits_{1}^{x} \frac{\log_e t}{1+t} \, dt

Then,

f(e)=\int\limits_{1}^{e \log_e t} \frac{dt}{1+t}

. . . (i)

and

f\left(\frac{1}{e}\right)=\int\limits_{1}^{\frac{1}{e}} \frac{\log_e t}{1+t} \, dt

. . .(ii)

Let

t=\frac{1}{u},\quad dt=\frac{-1}{u^{2}}\,du

and put in Eq. (ii), we get

f\left(\frac{1}{e}\right)=\int\limits_{1}^{e} \frac{\log(1/u)}{1+\frac{1}{u}} \cdot \frac{-1}{u^{2}}\,du = \int\limits_{1}^{e} \frac{\log u}{u(u+1)}\,du

Using change of variable

f\left(\frac{1}{e}\right)=\int\limits_{1}^{e} \frac{\log t}{t(t+1)}\,dt

. . . (iii)

From Eqs. (i) and (iii), we get

f(e)+f\left(\frac{1}{e}\right)=\int\limits_{1}^{e} \frac{\log t}{1+t}\,dt + \int\limits_{1}^{e} \frac{\log t}{t(1+t)}\,dt = \int\limits_{1}^{e \log t} \frac{gt}{t}

Take

\log t = v

, then

\frac{1}{t}\,dt = dv

f(e)+f\left(\frac{1}{e}\right)=\int\limits_{0}^{1} v\,dv = \left[\frac{v^{2}}{2}\right]_{0}^{1} = \frac{1}{2}

\therefore \quad f(e)+f\left(\frac{1}{e}\right)=\frac{1}{2}

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