Question

If $I_1={\int }_1^2\frac{dx}{\sqrt{1+x^2}}$ and $I_2={\int }_1^2\frac{dx}{x}$ then
compare value of $I_1$ and $I_2$

Answer 1

$I_1={\int }_1^2\frac{dx}{\sqrt{1+x^2}}$

We know that $\int \frac{dx}{\sqrt{a^2+x^2}}=\ln x+\sqrt{x^2+a^2}$

$={\left[\ln [x+\sqrt{x^2+1}]\right]}_1^2$

$=[\ln [2+\sqrt{2^2+1}]-\ln [1+\sqrt{1^2+1}]]$

$=[\ln [2+\sqrt{5}]-\ln [1+\sqrt{3}]]$

$=\ln \left[\frac{2+\sqrt{5}}{1+\sqrt{3}}\right]=\ln \frac{4.236}{2.732}=\ln 1.5505$

$I_2={\int }_1^2\frac{dx}{x}$

$={\left[\ln x\right]}_1^2$

$=[\ln 2-\ln 1]=\log \frac{2}{1}=\ln 2$

We have $\ln 1.5505<\ln 2$

Hence $I_1<I_2$