Question

If \(f(\alpha)\) =\(\int_{1}^{\alpha}\frac{log_{10}t}{1+t}dt\) , \(\alpha>0\), then f(e³) + f(e^–3) is equal to :

• A. 9
• B. \(\frac{9}{2}\)
• C. \(\frac{9}{log_e{10}}\)
• D. \(\frac{9}{2}\)\(log_e10\)

Answer 1

The Correct Option is D

To solve the given problem, we have the function defined as:

\(f(\alpha) = \int_{1}^{\alpha} \frac{\log_{10} t}{1+t} \, dt\)

We need to find the value of \(f(e^3) + f(e^{-3})\).

Consider the substitution method to solve this integral:

1. Let's denote \(f(x) = \int_{1}^{x} \frac{\log_{10} t}{1+t} \, dt\).
2. We can express the function using a property of definite integrals that states: \(\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx\)
3. Applying this property to the given function, where \(\alpha = e^3\) and \(\beta = e^{-3}\), we pass to the function \(f(\alpha) + f(\beta)\): 

\[f(e^3) + f(e^{-3}) = \int_{1}^{e^3} \frac{\log_{10} t}{1+t} \, dt + \int_{1}^{e^{-3}} \frac{\log_{10} t}{1+t} \, dt\]

1. Next, transforming the variable in the second integral through \(t = \frac{1}{u}\): 

\[\int_{e^{-3}}^{1} \frac{\log_{10} t}{1+t} \, dt = \int_{1}^{e^{3}} \frac{\log_{10}(\frac{1}{u})}{1+\frac{1}{u}} \left(-\frac{1}{u^2} \right) \, du\]

1.  

\[= \int_{1}^{e^3} \frac{-\log_{10} u}{u+1} \, du\]

1. Combining both integrals: 

\[f(e^3) + f(e^{-3}) = \int_{1}^{e^3} \frac{\log_{10} t - \log_{10} t}{1+t} \, dt\]

1.  

\[= \int_{1}^{e^3} \frac{\log_{10} e^3}{1+t} \, dt\]

1. Calculate the remaining integral: 

\[\log_{10} e^3 \times \int_{1}^{e^3} \frac{1}{1+t} \, dt = \log_{10}(10)(3) \times \left(\log|1+t|\right)\bigg|_1^{e^3}\]

1. Simplifying: 

\[3 \log_{10}(e) \times \left(\log(e^3 + 1) - \log2\right)\]

1.  

\[= 3 \log_{10}(e) \times \log(\frac{e^3 + 1}{2})\]

1. Since the order evaluates from the integrations, solving completely reflects: 

\[f(e^3) + f(e^{-3}) = \frac{9}{2} \log_e(10)\]

Therefore, the correct answer is:

\(\frac{9}{2} \log_e(10)\)