AP EAMCET · Maths · Definite Integration
\(\int_1^{e^2} \frac{d x}{x(1+\log x)^2}=\)
- A \(\frac{2}{3}\)
- B \(\frac{1}{3}\)
- C \(\frac{3}{2}\)
- D \(\log 2\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
\(I=\int_1^{e^2} \frac{d x}{x(I+\log x)^2}\) Let \(\quad 1+\log x=t\) \(\frac{1}{x} d x=d t\) \(\Rightarrow \quad I=\int_1^3 \frac{d t}{t^2}=\left[\frac{t^{-1}}{-1}\right]_1^3=\left(\frac{3^{-1}}{-1}+\frac{1^{-1}}{1}\right)\) \(I=-\frac{1}{3}+1=\frac{2}{3}\)
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