Question

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

Answer: (A)

Putting $t=\frac{1}{u}$ in $I_1$, we get
$I_1=\underset{1/x}{\overset{1}{\int }}\frac{-\frac{1}{u^2}du}{1+\frac{1}{u^2}}=-\underset{1/x}{\overset{1}{\int }}\frac{du}{1+u^2}$
$=\underset{1}{\overset{1/x}{\int }}\frac{du}{1+u^2}=-\underset{1}{\overset{1/x}{\int }}\frac{dt}{1+t^2}=I_2$
$\therefore I_1=I_2$