JEE Mains · Maths · STD 12 - 7.2 definite integral
If \(f (\alpha)=\int_{1}^{\alpha} \frac{\log _{10} t}{1+t} d t, \alpha>0\), then \(f \left( e ^{3}\right)+ f \left( e ^{-3}\right)\) is equal to.
- A \(9\)
- B \(\frac{9}{2}\)
- C \(\frac{9}{\log _{e}(10)}\)
- D \(\frac{9}{2 \log _{e}(10)}\)
Answer & Solution
Correct Answer
(D) \(\frac{9}{2 \log _{e}(10)}\)
Step-by-step Solution
Detailed explanation
\(f \left( e ^{3}\right)=\int_{1}^{ e ^{3}} \frac{\ell nt }{\ell n 10(1+ t )} dt \ldots \ldots(1)\) \(f (\alpha)=\int_{1}^{\alpha} \frac{\ell n t }{(\ell \operatorname{nn} 10)(1+ t )} dt\) \(t =\frac{1}{ x } \Rightarrow x =\frac{1}{ t }\) \(dt =\frac{-1}{ x ^{2}} dx\)…
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