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Let $I_1=\int_1^2 \frac{d x}{\sqrt{1+x^2}}$ and $I_2=\int_1^2 \frac{d x}{x}$ then
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The correct answer is:
$I_2 \gt I_1$
We have, $\left(1+x^2\right) \gt x^2, \forall x ; \sqrt{1+x^2} \gt x, \forall x \in(1,2)$
$\begin{aligned}
& \Rightarrow \frac{1}{\sqrt{1+x^2}} \lt \frac{1}{x}, \forall x \in(1,2) \Rightarrow \int_1^2 \frac{d x}{\sqrt{1+x^2}} \lt \int_1^2 \frac{d x}{x} \\
& \Rightarrow I_1 \lt I_2 \Rightarrow I_2 \gt I_1 .
\end{aligned}$
$\begin{aligned}
& \Rightarrow \frac{1}{\sqrt{1+x^2}} \lt \frac{1}{x}, \forall x \in(1,2) \Rightarrow \int_1^2 \frac{d x}{\sqrt{1+x^2}} \lt \int_1^2 \frac{d x}{x} \\
& \Rightarrow I_1 \lt I_2 \Rightarrow I_2 \gt I_1 .
\end{aligned}$
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