44.3k views
2 votes
How would I find the integral of
\int(tdt)/(t^4+2)?

User Florex
by
8.5k points

1 Answer

7 votes
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
User Mohd Shahril
by
8.4k points

Related questions

2 answers
0 votes
134k views
1 answer
2 votes
6.3k views
1 answer
2 votes
200k views