Question

How would I find the integral of
`\int(tdt)/(t^4+2)`?

Answer 1

Let
`t=\sqrt y`, so that
`t^2=y`,
`t^4=y^2`, and
`\mathrm dt=(\mathrm dy)/(2\sqrt y)`. Then


`\displaystyle\int\frac t{t^4+2}\,\mathrm dt=\int(\sqrt y)/(2\sqrt y(y^2+2))\,\mathrm dy=\frac12\int(\mathrm dy)/(y^2+2)`

Now let
`y=\sqrt2\tan z`, so that
`\mathrm dy=\sqrt2\sec^2z\,\mathrm dz`. Then


`\displaystyle\frac12\int(\mathrm dy)/(y^2+2)=\frac12\int(\sqrt2\sec^2z)/((\sqrt2\tan z)+2)\,\mathrm dz=\frac{\sqrt2}4\int(\sec^2z)/(\tan^2z+1)\,\mathrm dz=\frac1{2\sqrt2}\int\mathrm dz=\frac1{2\sqrt2}z+C`

Transform back to
`y` to get


`\frac1{2\sqrt2}\arctan\left(\frac y{\sqrt2}\right)+C`

and again to get back a result in terms of
`t`.


`\frac1{2\sqrt2}\arctan\left((t^2)/(\sqrt2)\right)+C`