Problem source

Show, by means of a suitable change of variable,
or otherwise, that
\[
\int_{0}^{\infty}\mathrm{f}((x^{2}+1)^{1/2}+x)\,{\mathrm d}x
=\frac{1}{2}
\int_{1}^{\infty}(1+t^{-2})\mathrm{f}(t)\,{\mathrm d}t.
\]
Hence, or otherwise, show that
\[
\int_{0}^{\infty}((x^{2}+1)^{1/2}+x)^{-3}\,{\mathrm d}x
=\frac{3}{8}.
\]

Solution source

\begin{align*}
&& t &= (x^2+1)^{1/2}+x \\
&& 1&=t^2-2tx \\
&& x &= \frac{t^2-1}{2t} = \frac12 \left (t - \frac1t\right) \\
&& \frac{\d x}{\d t} &= \frac12 \left ( 1+ \frac{1}{t^2} \right) \\
\Rightarrow && \int_0^{\infty} f((x^2+1)^{1/2}+x) \d x &= \int_{t=1}^{t = \infty}f(t) \frac12(1 + t^{-2}) \d t\\
&&&= \frac12 \int_1^{\infty}(1+t^{-2})f(t) \d t
\end{align*}

\begin{align*}
\int_{0}^{\infty}((x^{2}+1)^{1/2}+x)^{-3}\,{\mathrm d}x &= \frac12 \int_1^{\infty}(1+t^{-2})t^{-3} \d t \\
&= \frac12 \left [\frac{-1}{2}t^{-2}-\frac{1}{4}t^{-4} \right]_{1}^{\infty} \\
&= \frac12 \left ( \frac12 + \frac14\right) = \frac38
\end{align*}