Question

Let $I_1={\int }_1^2\frac{1}{\sqrt{1+x^2}}dx$ and $I_2={\int }_1^2\frac{1}{x}dx$. Then

• A. $I_1>I_2$
• B. $I_2>I_1$
• C. $I_1=I_2$
• D. $I_1>2I_2$

Answer 1

Answer: (B)

$I_2>I_1$

$I_1={\int }_1^2\frac{1}{\sqrt{1+x^2}}dx$
$={\left(\log (x+\sqrt{1+x^2})\right)}_1^2$
$=\log (2+\sqrt{5})-\log (1+\sqrt{2})=\log \left(\frac{(2+\sqrt{5})}{(1+\sqrt{2})}\right)$
$I_2={\int }_1^2\frac{1}{x}dx={\left(\log x\right)}_1^2=\log 2-\log 1=\log 2$
$\therefore I_2>I_1$