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  4. If I1= displaystyle∫x1(dt/1+t2) and I2= displaystyle∫11/x(dt/1+t2) for x > 0, then

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Q. If $I_1=\displaystyle\int_{x}^{1}\frac{dt}{1+t^2}$ and $I_2=\displaystyle\int_{1}^{1/x}\frac{dt}{1+t^2}$ for $x > 0,$ then

Integrals

Solution:

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