If $\mathrm{f}(\alpha)=\int_{1}^{\alpha} \frac{\log _{10} t}{1+t} d t, \alpha>0$, then $\mathrm{f}\left(\mathrm{e}^{3}\right)+\mathrm{f}\left(\mathrm{e}^{-3}\right)$ is equal to :

(D) $\frac{9}{2 \log _{e}(10)}$

(C) $\frac{9}{\log _{e}(10)}$

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Definite Integrals

f (α) = \int_{1}^{α} \frac{l o g _{10} t}{1 + t} d t, α > 0

, then

f (e^{3}) + f (e^{- 3})

is equal to :

Answer: (D)

Write

lo g_{10} t = \frac{l n t}{l n 10}

. For

f (α)

, substitute

t = 1/ x

to get

f (α) = \frac{1}{l n 10} \int_{1}^{1/ α} \frac{l n x}{x ( x + 1 )} d x

. Therefore

f (e^{- 3}) = \frac{1}{l n 10} \int_{1}^{e^{3}} \frac{l n t}{t ( t + 1 )} d t

.

Adding

f (e^{3})

and

f (e^{- 3})

f (e^{3}) + f (e^{- 3}) = \frac{1}{ln 10} \int_{1}^{e^{3}} \frac{ln t}{1 + t} (1 + \frac{1}{t}) d t = \frac{1}{ln 10} \int_{1}^{e^{3}} \frac{ln t}{t} d t

Let

u = ln t

, so

d u = \frac{d t}{t}

. When

t = 1

u = 0

; when

t = e^{3}

u = 3

. Thus:

\frac{1}{ln 10} \int_{0}^{3} u d u = \frac{1}{ln 10} \cdot \frac{9}{2} = \frac{9}{2 lo g _{e} 10}

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option