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$\int_{1}^{28} \frac{\mathrm{d} x}{x(1+\log x)^{2}}=$
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Verified Answer
The correct answer is:
$\frac{\log 2}{(1+\log 2)}$
$I=\int_{1}^{2} \frac{d x}{x(1+\log x)}$
Put $\quad 1+\log x=t \Rightarrow \frac{1}{x} d x=d t$
When $x=1, t=1$ and when $x=2, t=1+\log 2$
$\begin{aligned} I &=\int_{1}^{1+\log 2} \frac{d t}{t}=[\log t]_{1}^{1+\log 2}=\log (1+\log 2)-\log 1 \\ &=\log (1+\log 2) \end{aligned}$
Put $\quad 1+\log x=t \Rightarrow \frac{1}{x} d x=d t$
When $x=1, t=1$ and when $x=2, t=1+\log 2$
$\begin{aligned} I &=\int_{1}^{1+\log 2} \frac{d t}{t}=[\log t]_{1}^{1+\log 2}=\log (1+\log 2)-\log 1 \\ &=\log (1+\log 2) \end{aligned}$
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