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\(\int_1^{e^2} \frac{d x}{x(1+\log x)^2}=\)
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Verified Answer
The correct answer is:
\(\frac{2}{3}\)
\(I=\int_1^{e^2} \frac{d x}{x(I+\log x)^2}\)
Let \(\quad 1+\log x=t\)
\(\frac{1}{x} d x=d t\)
\(\Rightarrow \quad I=\int_1^3 \frac{d t}{t^2}=\left[\frac{t^{-1}}{-1}\right]_1^3=\left(\frac{3^{-1}}{-1}+\frac{1^{-1}}{1}\right)\)
\(I=-\frac{1}{3}+1=\frac{2}{3}\)
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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