Question

If $I_1={\int }_x^1\frac{1}{1+t^2}dt$ and $I_2={\int }_1^{1/x}\frac{1}{1+t^2}dt$ for $x>0$, then

• A. $I_1=I_2$
• B. $I_1>I_2$
• C. $I_2>I_1$
• D. none of these

Answer 1

The correct option is A $I_1=I_2$

$I_1={\int }_x^1\frac{1}{1+t^2}dt$
Substitute $t=\frac{1}{u}\Rightarrow dt=-\frac{1}{u^2}du$
We get
$I_1={\int }_{1/x}^1\frac{1}{1+\frac{1}{u^2}}(-\frac{1}{u^2})du=-{\int }_{1/x}^1\frac{du}{u^2+1}={\int }_1^{1/x}\frac{1}{1+u^2}du=I_2$