Show that $\int_x^1\frac{dt}{1+t^2}=\int_1^{1/x}\frac{dt}{1+t^2},\;x>0.$

Show that $$\int_x^1\frac{dt}{1+t^2}=\int_1^{1/x}\frac{dt}{1+t^2},\;x>0.$$

So, I'm learning about integration techniques, and I get this exercise. We've been practicing $u$-substitution, and I think that might be involved here somehow, but I don't understand what I need to do to proceed. Thank you for your help!

• Your bounds on the second integral are in the wrong order: the integral on the left is positive and that on the right is negative. Feb 5, 2014 at 1:37
\begin{align} u & = \frac 1 t \\[8pt] du & = \frac{-dt}{t^2} \\[8pt] -t^2\,du & = dt \\[8pt] \frac{-du}{u^2} & = dt \end{align} When $t=x$ the $u=1/x$.
So $$\int_1^x \frac{dt}{1+t^2} = \int_1^{1/x} \frac{-du/u^2}{1+(1/u)^2} = \int_1^{1/x} \frac{-du}{u^2+1}.$$
the substitution $t \leftarrow 1/t$ yields $$\int_1^x {dt\over 1 + t^2 } = \int_{1/x}^1 {dt/t^2\over 1 + (1/t)^2} = \int_{1/x}^1 {dt\over 1 + t^2}.$$