Let $\mathrm{I}_1=\int_1^2 \frac{1}{\sqrt{1+\mathrm{x}^2}} \mathrm{dx}$ and $\mathrm{I}_2=\int_1^2 \frac{1}{x} d x$, then

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Let $\mathrm{I}_1=\int_1^2 \frac{1}{\sqrt{1+\mathrm{x}^2}} \mathrm{dx}$ and $\mathrm{I}_2=\int_1^2 \frac{1}{x} d x$, then, $\mathrm{I}_1 > \mathrm{I}_2$, $\mathrm{I}_2 > \mathrm{I}_1$, $\mathrm{I}_1=\mathrm{I}_2$, None of these

MARKS App · Accepted Answer

If $I_1=\int_1^2 \frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}^2}}, I_2=\int_1^2 \frac{\mathrm{dx}}{\mathrm{x}}$ $I_1=\ell n \left(\frac{2+\sqrt{5}}{1+\sqrt{2}}\right), \quad I_2=\ell n 2 \Rightarrow I_1 < I_2$

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