Indefinite integral (04)

[Grigorios Kostakos]

Evaluate \[\displaystyle\int{\frac{x^2-1}{({x^2+1})\,\sqrt{1+x^4}}\,dx}\,.\]

[jacks]

\(\displaystyle\int{\frac{x^2-1}{({x^2+1})\,\sqrt{1+x^4}}\,dx}=\int{\frac{1-\frac{1}{x^2}}{\bigl({x+\frac{1}{x}}\bigr)\,\sqrt{\frac{1}{x^2}+x^2}}\,dx}=\)

\(\displaystyle\int{\frac{1-\frac{1}{x^2}}{\bigl({x+\frac{1}{x}}\bigr)\,\sqrt{\bigl({x+\frac{1}{x}}\bigr)^2-2}}\,dx}\,\mathop{=\!=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,x+\frac{1}{x}} \\
{dt\,=\,({1-\frac{1}{x^2}})\,dx} \\
\end{subarray}}\,\int{\frac{1}{t\,\sqrt{t^2-2}}\,dt}\,\mathop{=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{u^2\,=\,t^2-2} \\
{dt\,=\,\frac{u}{t}\,du} \\
\end{subarray}}\)

\(\displaystyle\int{\frac{1}{u^2+\bigl({\sqrt2\,}\bigr)^2}\,du}=\dfrac{1}{\sqrt{2}}\,\tan^{-1}\bigl({\tfrac{u}{\sqrt2}}\bigr)+c\)

So \(\displaystyle\int{\frac{x^2-1}{({x^2+1})\,\sqrt{1+x^4}}\,dx}=\dfrac{1}{\sqrt{2}}\,\tan^{-1}\biggl({\tfrac{\sqrt{\frac{1}{x^2}+x^2}}{\sqrt2}}\biggr)+c\,.\)

[Grigorios Kostakos]

2nd solution: \(\displaystyle\int{\frac{x^2-1}{({x^2+1})\,\sqrt{1+x^4}}\,dx}=\int{\frac{({x^2-1})({x^2+1})}{({x^2+1})^2\,\sqrt{1+x^4}}\,dx}=\)

\(\displaystyle\int{\frac{x^4-1}{({x^4+2x^2+1})\,\sqrt{1+x^4}}\,dx}=\int{\frac{x^4-1}{2x^2\,\bigl({1+\frac{x^4+1}{2x^2}}\bigr)\,\sqrt{1+x^4}}\,dx}=\)

\(\displaystyle\int{\frac{x^4-1}{\Bigl({1+{\sqrt{\frac{x^4+1}{2x^2}}\,}^2\,}\Bigr)\,2x^2\,\sqrt{1+x^4}}\,dx}\,\mathop{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,\sqrt{\frac{x^4+1}{2x^2}}} \\
{\frac{\sqrt{2}}{2}\,dt\,=\,\frac{x^4-1}{2x^2\,\sqrt{1+x^4}}\,dx} \\
\end{subarray}}\,\frac{\sqrt{2}}{2}\,\int{\frac{1}{1+t^2}\,dt}=\)

\(\displaystyle\frac{\sqrt{2}}{2}\,\arctan{t}+c\,\mathop{=\!=\!=\!=\!=\!=}\limits^{t\,=\,\sqrt{\frac{x^4+1}{2x^2}}}\,\frac{\sqrt{2}}{2}\,\arctan\Bigl({\sqrt{\tfrac{x^4+1}{2x^2}}\,}\Bigr)+c=\frac{\sqrt{2}}{2}\,\arctan\Bigl({\tfrac{\sqrt{2}}{2}\sqrt{x^2+\tfrac{1}{x^2}}\,}\Bigr)+c\,.\)