If f(α)=∫1α log10⁡t1+tdt,α>0, then fe3+fe−3 is equal to

f(α)=∫1α log10⁡t1+tdt…(i); fe3=∫1e3 log10⁡t1+tdt ....(ii)Put t=1x in (i), we getf(α)=∫11/α log10⁡(1/x)1+(1/x)−1x2dx=∫11/α log10⁡xx(x+1)dx∴fe−3=∫1e3 log10⁡xx(x+1)dx=∫1e3 log10⁡tt(t+1)dt ....iiiAdding (ii) and (iii) , we get fe3+fe−3=∫1e3 log10⁡tt+11+1tdt=∫1e3 log10⁡ttdtPut log10⁡t=z⇒1loge⁡10⋅1tdt=dzt=1⇒z=0,t=e3⇒z=3/loge⁡10∴fe3+fe−3=∫03/loge⁡10 z⋅loge⁡10dz=loge⁡10⋅z2203/loge⁡10=loge⁡1029loge⁡102−0=92loge⁡10

If f(α)=∫1α log10⁡t1+tdt,α>0, then fe3+fe−3 is equal to

f(α)=∫1α log10⁡t1+tdt…(i); fe3=∫1e3 log10⁡t1+tdt ....(ii)Put t=1x in (i), we getf(α)=∫11/α log10⁡(1/x)1+(1/x)−1x2dx=∫11/α log10⁡xx(x+1)dx∴fe−3=∫1e3 log10⁡xx(x+1)dx=∫1e3 log10⁡tt(t+1)dt ....iiiAdding (ii) and (iii) , we get fe3+fe−3=∫1e3 log10⁡tt+11+1tdt=∫1e3 log10⁡ttdtPut log10⁡t=z⇒1loge⁡10⋅1tdt=dzt=1⇒z=0,t=e3⇒z=3/loge⁡10∴fe3+fe−3=∫03/loge⁡10 z⋅loge⁡10dz=loge⁡10⋅z2203/loge⁡10=loge⁡1029loge⁡102−0=92loge⁡10

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If $f (α) = \int_{1}^{α} \frac{\log_{10} t}{1 + t} dt, α > 0, then f (e^{3}) + f (e^{- 3})$ is equal to

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$\frac{9}{2}$

$\frac{9}{2 \log_{e} (10)}$

$\frac{9}{\log_{e} (10)}$

answer is D.

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Detailed Solution

$f (α) = \int_{1}^{α} \frac{\log_{10} t}{1 + t} dt \dots (i); f (e^{3}) = \int_{1}^{e^{3}} \frac{\log_{10} t}{1 + t} dt . . . . (ii)$
Put $t = \frac{1}{x} in (i)$ , we get
$f (α) = \int_{1}^{1 / α} \frac{\log_{10} (1 / x)}{1 + (1 / x)} (\frac{- 1}{x^{2}}) dx = \int_{1}^{1 / α} \frac{\log_{10} x}{x (x + 1)} dx$
$∴ f (e^{- 3}) = \int_{1}^{e^{3}} \frac{\log_{10} x}{x (x + 1)} dx = \int_{1}^{e^{3}} \frac{\log_{10} t}{t (t + 1)} dt . . . . (iii)$
Adding (ii) and (iii) , we get
$f (e^{3}) + f (e^{- 3}) = \int_{1}^{e^{3}} \frac{\log_{10} t}{t + 1} (1 + \frac{1}{t}) dt = \int_{1}^{e^{3}} \frac{\log_{10} t}{t} dt$
Put $\log_{10} t = z \Rightarrow \frac{1}{\log_{e} 10} \cdot \frac{1}{t} dt = dz$
$t = 1 \Rightarrow z = 0, t = e^{3} \Rightarrow z = 3 / \log_{e} 10$
$\begin{array}{l} ∴ f (e^{3}) + f (e^{- 3}) = \int_{0}^{3 / \log_{e} 10} z \cdot \log_{e} 10 dz \\ = \log_{e} 10 \cdot {[\frac{z^{2}}{2}]}_{0}^{3 / \log_{e} 10} = \frac{\log_{e} 10}{2} (\frac{9}{{(\log_{e} 10)}^{2}} - 0) \\ = \frac{9}{2 \log_{e} 10} \end{array}$