definite integration
Posted: Thu Jul 07, 2016 2:51 pm
\(\displaystyle \int_{0}^{\infty}\frac{1}{(x+\sqrt{x^2+1})^3}dx\)
Substituting \(x=\sinh{t}\) we have \(\sqrt{x^2+1}=\cosh{t}\) and \(dx=\cosh{t}\,dt\). So
\[J=\displaystyle \int_{0}^{\infty}\frac{1}{\bigl(x+\sqrt{x^2+1}\,\bigr)^3}\,dx=\int_{0}^{\infty}\frac{\cosh{t}}{(\sinh{t}+\cosh{t})^3}\,dt\,.\]
But \(\cosh{t}=\dfrac{e^t+e^{-t}}{2}\) and \((\sinh{t}+\cosh{t})^3=e^{3t}\) and so \begin{align*}
J&=\displaystyle\int_{0}^{\infty}\frac{e^t+e^{-t}}{2e^{3t}}\,dt=\frac{1}{2}\int_{0}^{\infty}e^{-4t}\,dt+\frac{1}{2}\int_{0}^{\infty}{e^{-2t}}\,dt\,\mathop{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{u\,=e^{-4t}\,,\;-\frac{1}{4}\,du=dt} \\
{v\,=e^{-2t}\,,\;-\frac{1}{2}\,dv=dt}
\end{subarray}}-\frac{1}{2}\,\frac{1}{4}\int_{1}^{0}u\,du-\frac{1}{2}\frac{1}{2}\int_{1}^{0}{v}\,dv\\
& =\frac{1}{2}\,\frac{1}{4}\int_{0}^{1}u\,du+\frac{1}{2}\frac{1}{2}\int_{0}^{1}{v}\,dv=\frac{1}{2}\,\frac{1}{4}\,1+\frac{1}{2}\,\frac{1}{2}\,1=\frac{3}{8}\,.
\end{align*}