Problem source

Explain why the use of the substitution $x=\dfrac{1}{t}$ does not demonstrate that the integrals 
\[
\int_{-1}^{1}\frac{1}{(1+x^{2})^{2}}\,\mathrm{d}x\quad\mbox{ and }\quad\int_{-1}^{1}\frac{-t^{2}}{(1+t^{2})^{2}}\,\mathrm{d}t
\]
are equal. 
Evaluate both integrals correctly.

Solution source

When we apply the substitution $x = \frac1{t}$, $t$ runs from $-1 \to -\infty$ as $x$ goes from $-1 \to 0$. Then it runs from $\infty \to 1$ as $x$ runs from $0 \to 1$. So we would be able to show that:

\[ \int_{-1}^{1}\frac{1}{(1+x^{2})^{2}}\,\mathrm{d}x = \int_{-1}^{-\infty}\frac{-t^{2}}{(1+t^{2})^{2}}\,\mathrm{d}t + \int_{\infty}^1 \frac{-t^{2}}{(1+t^{2})^{2}}\,\mathrm{d}t \]

Let $x = \tan u, \d x = \sec^2 u \d u$
\begin{align*}
\int_{-1}^1 \frac1{(1+x^2)^2} \d x &= \int_{u = -\pi/4}^{u = \pi/4} \frac{\sec^2 u}{(1+\tan^2 u)^2} \d u \\
&=  \int_{u = -\pi/4}^{u = \pi/4} \frac{1}{\sec^2 u} \d u \\
&= \int_{-\pi/4}^{\pi/4} \cos^2 u \d u \\
&= \int_{-\pi/4}^{\pi/4} \frac{1 + \cos 2 u}{2} \d u \\
&= \left [ \frac{2u + \sin 2u}{4} \right]_{-\pi/4}^{\pi/4} \\
&= \frac{\pi}{4} + \frac{1}{2} 
\end{align*}

Let $t = \tan u, \d t = \sec^2 u \d u$
\begin{align*}
\int_{-1}^1 \frac{-t^2}{(1+t^2)^2} \d x &= \int_{u = -\pi/4}^{u = \pi/4} \frac{-\tan^2 u \sec^2 u}{(1+\tan^2 u)^2} \d u \\
&=  -\int_{u = -\pi/4}^{u = \pi/4} \frac{\tan^2 u}{\sec^2 u} \d u \\
&= -\int_{-\pi/4}^{\pi/4} \sin^2 u \d u \\
&= -\int_{-\pi/4}^{\pi/4} \frac{1 - \cos 2 u}{2} \d u \\
&= -\left [ \frac{2u - \sin 2u}{4} \right]_{-\pi/4}^{\pi/4} \\
&= \frac{1}{2}-\frac{\pi}{4}  
\end{align*}