MHT CET · Maths · Definite Integration
\(\int_{1}^{28} \frac{\mathrm{d} x}{x(1+\log x)^{2}}=\)
- A \(\cdot \log 2\)
- B \(1+\log 2\)
- C \(\frac{\log 2}{(1+\log 2)}\)
- D \(\frac{1}{(1+\log 2)}\)
Answer & Solution
Correct Answer
(C) \(\frac{\log 2}{(1+\log 2)}\)
Step-by-step Solution
Detailed explanation
\(I=\int_{1}^{2} \frac{d x}{x(1+\log x)}\)
Put \(\quad 1+\log x=t \Rightarrow \frac{1}{x} d x=d t\)
When \(x=1, t=1\) and when \(x=2, t=1+\log 2\)
\(I =\int_{1}^{1+\log 2} \frac{d t}{t}=[\log t]_{1}^{1+\log 2}=\log (1+\log 2)~-\) \(\log 1\)
\(=\log (1+\log 2)\)
Put \(\quad 1+\log x=t \Rightarrow \frac{1}{x} d x=d t\)
When \(x=1, t=1\) and when \(x=2, t=1+\log 2\)
\(I =\int_{1}^{1+\log 2} \frac{d t}{t}=[\log t]_{1}^{1+\log 2}=\log (1+\log 2)~-\) \(\log 1\)
\(=\log (1+\log 2)\)
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