Let \(t=\sqrt y\), so that \(t^2=y\), \(t^4=y^2\), and \(\mathrm dt=\dfrac{\mathrm dy}{2\sqrt y}\). Then
\(\displaystyle\int\frac t{t^4+2}\,\mathrm dt=\int\frac{\sqrt y}{2\sqrt y(y^2+2)}\,\mathrm dy=\frac12\int\frac{\mathrm dy}{y^2+2}\)
Now let \(y=\sqrt2\tan z\), so that \(\mathrm dy=\sqrt2\sec^2z\,\mathrm dz\). Then
\(\displaystyle\frac12\int\frac{\mathrm dy}{y^2+2}=\frac12\int\frac{\sqrt2\sec^2z}{(\sqrt2\tan z)+2}\,\mathrm dz=\frac{\sqrt2}4\int\frac{\sec^2z}{\tan^2z+1}\,\mathrm dz=\frac1{2\sqrt2}\int\mathrm dz=\dfrac1{2\sqrt2}z+C\)
Transform back to \(y\) to get
\(\dfrac1{2\sqrt2}\arctan\left(\dfrac y{\sqrt2}\right)+C\)
and again to get back a result in terms of \(t\).
\(\dfrac1{2\sqrt2}\arctan\left(\dfrac{t^2}{\sqrt2}\right)+C\)
