Let I=∫12dx1+x2 and J=∫12dxx, then

In[1, 2], x2 1x2+1 ⇒1x>1x2+1⇒I<J

Let I=∫12dx1+x2 and J=∫12dxx, then

In[1, 2], x2 1x2+1 ⇒1x>1x2+1⇒I<J

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Let $I = \int_{1}^{2} \frac{d x}{\sqrt{1 + x^{2}}}$ and $J = \int_{1}^{2} \frac{d x}{x},$ then

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$I < J$

$I = J$

None of these

$I > J$

answer is B.

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Detailed Solution

In $[1, 2],$ $x^{2} < x^{2} + 1 \Rightarrow \sqrt{x^{2}} < \sqrt{x^{2} + 1}$

$\Rightarrow \frac{1}{\sqrt{x^{2}}} > \frac{1}{\sqrt{x^{2} + 1}}$

$\Rightarrow \frac{1}{x} > \frac{1}{\sqrt{x^{2} + 1}} \Rightarrow I < J$

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Let I=∫12dx1+x2 and J=∫12dxx, then

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Let $I = \int_{1}^{2} \frac{d x}{\sqrt{1 + x^{2}}}$ and $J = \int_{1}^{2} \frac{d x}{x},$ then