If f(α) = ∫ _ 1 ^ α _ 10 t 1 + t dt, α > 0 , then f( e ^ 3 ) + f( e...

Given,

$f\left(\alpha\right) = {∫}_{1}^{\alpha}\dfrac{\log_{10}t}{1 + t}dt$

So, $f\left({e}^{3}\right) = {∫}_{1}^{{e}^{3}}\dfrac{\ln t}{\ln 10\left(1 + t\right)}dt ⋯ \left(1\right)$

Again $f\left(\alpha\right) = {∫}_{1}^{\alpha}\dfrac{\ln t}{\left(\ln 10\right)\left(1 + t\right)}dt$

Now let $t = \dfrac{1}{x} ⇒ x = \dfrac{1}{t}⇒ dt = \dfrac{- 1}{{x}^{2}}dx$

So, $f\left(\alpha\right) = {∫}_{1}^{\dfrac{1}{a}}\dfrac{- lnnx}{\left(\ln 10\right)\left(1 + \dfrac{1}{x}\right)}\left(- \dfrac{1}{{x}^{2}}\right)dx$

$⇒ f\left(\alpha\right) = \dfrac{1}{\ln 10}{∫}_{1}^{\dfrac{1}{\alpha}}\dfrac{\ln x}{x\left(x + 1\right)}dx$

So, $f\left({e}^{- 3}\right) = \dfrac{1}{\ln 10}{∫}_{1}^{{e}^{3}}\dfrac{\ln t}{t\left(t + 1\right)}dt \ldots \left(2\right)$

Now adding equation \left(1\right)$and \left(2\right)$ we get,

$f\left({e}^{3}\right) + f\left({e}^{- 3}\right)= \left(\dfrac{1}{lnn10}\right){∫}_{1}^{{c}^{3}}\dfrac{\ln t}{\left(1 + t\right)}\left[1 + \dfrac{1}{t}\right]dt$

$⇒ f\left({e}^{3}\right) + f\left({e}^{- 3}\right) = \left(\dfrac{1}{\ln 10}\right){∫}_{1}^{3}\dfrac{\ln t}{t}dt$

Now let $\ln t = r⇒ \dfrac{dt}{t} = dr$

So, $f\left({e}^{3}\right) + f\left({e}^{- 3}\right) = \dfrac{1}{\ln 10}{∫}_{0}^{3}rdr$

$⇒ f\left({e}^{3}\right) + f\left({e}^{- 3}\right){= \left(\dfrac{1}\ln10\right)\left(\dfrac{{r}^{2}}{2}\right)}_{0}^{3}$

$⇒ f\left({e}^{3}\right) + f\left({e}^{- 3}\right) = \left(\dfrac{1}{\log 10}\right)\left(\dfrac{9}{2}\right)$

$⇒ f\left({e}^{3}\right) + f\left({e}^{- 3}\right) = \dfrac{9}{2\log_{e}10}$