Prove that
![\[ \int_x^1 \frac{dt}{1+t^2} = \int_1^{\frac{1}{x}} \frac{dt}{1+t^2} \qquad \text{for } x > 0. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-00367c40e653407dd964b26f8eade4dd_l3.png "Rendered by QuickLaTeX.com")
Let
![\[ u = g(t) = \frac{1}{t}, \qquad du = -\frac{1}{t^2} \, dt. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f64653d1324ea9e302f65ccda9166e65_l3.png "Rendered by QuickLaTeX.com")
These equations imply,
![\[ t = \frac{1}{u}, \qquad dt = -t^2 \,du = -\frac{1}{u^2} \, du. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3074068134a556048b825e463ac05e73_l3.png "Rendered by QuickLaTeX.com")
Therefore, substituting for 
we have,
![\begin{align*} \int_x^1 \frac{dt}{1+t^2} &= -\int_{g(x)}^{g(1)} \frac{\frac{1}{u^2} \, du}{1+\frac{1}{u^2}} \\[10pt] &= \int_{g(1)}^{g(x)} \frac{\frac{1}{u^2} \, du}{1 + \frac{1}{u^2}} \\[10pt] &= \int_1^{\frac{1}{x}} \frac{du}{1+u^2} \\[10pt] &= \int_1^{\frac{1}{x}} \frac{dt}{1+t^2}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ea92db33e8ee1e53d3e9c9d37da6e0b2_l3.png "Rendered by QuickLaTeX.com")
The final step follows since we can rename the variable of integration (since 
since all we’ve done is rename the variable)
Prove that
Let
These equations imply,
Therefore, substituting for we have,
The final step follows since we can rename the variable of integration (since since all we’ve done is rename the variable)