Question

How would I find the integral of ="\int\frac{tdt}{t^4+2}" alt="\int\frac{tdt}{t^4+2}" align="absmiddle" class="latex-formula">?

Answer 1

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 $t$.

$\dfrac1{2\sqrt2}\arctan\left(\dfrac{t^2}{\sqrt2}\right)+C$