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