Question

lf $I_1={\int }_0^1\frac{1}{|x|}dx$ and $I_2={\int }_0^1\frac{1}{\sqrt{1+x^2}}dx$, then

• A. $I_1=I_2$
• B. $I_1<I_2$
• C. $I_1>I_2$
• D. $I_1=5I_2$

Answer 1

The correct option is C $I_1>I_2$

$I_1={\int }_0^1\frac{1}{\left|x\right|}dx$

$0<x<1\Rightarrow \left|x\right|=x>0$

$I_1={\int }_0^1\frac{1}{x}dx={\left[\ln x\right]}_0^1=0-\underset{x\to 0}{lim}\ln x$, which tends to $+\infty$

$I_2={\int }_0^1\frac{1}{\sqrt{1+x^2}}dx={\left[\ln |x+\sqrt{1+x^2}|\right]}_0^1=\ln (1+\sqrt{2})$

Clearly, $I_1>I_2$

Hence, option C.