## Calculus 8th Edition

$\frac{1}{2\sqrt{2}}\tan^{-1}\frac{t^2}{\sqrt{2}}+C$
$\int \frac{t}{t^4+2}dt$ Let $u=t^2$. Then $du=2t\ dt$, and $t\ dt=\frac{1}{2}du$. $=\int\frac{t}{(t^2)^2+2}dt$ $=\int\frac{1}{u^2+2}*\frac{1}{2}du$ $=\frac{1}{2}\int\frac{1}{u^2+2}du$ Use the trig substitution $u=\sqrt{2}\tan\theta$. Then $du=\sqrt{2}\sec^2\theta\ d\theta$. $=\frac{1}{2}\int\frac{1}{(\sqrt{2}\tan\theta)^2+2}*\sqrt{2}\sec^2\theta\ d\theta$ $=\frac{\sqrt{2}}{2}\int\frac{\sec^2\theta}{2\tan^2\theta+2}d\theta$ $=\frac{1}{\sqrt{2}}\int\frac{\sec^2\theta}{2(\tan^2\theta+1)}d\theta$ $=\frac{1}{\sqrt{2}}\int\frac{\sec^2\theta}{2\sec^2\theta}d\theta$ $=\frac{1}{\sqrt{2}}\int\frac{1}{2}d\theta$ $=\frac{1}{2\sqrt{2}}\theta+C$ Note that since $u=\sqrt{2}\tan\theta$, $\tan \theta=\frac{u}{\sqrt{2}}$. Then $\theta=\tan^{-1}\frac{u}{\sqrt{2}}$. $=\frac{1}{2\sqrt{2}}\tan^{-1}\frac{u}{\sqrt{2}}+C$ $=\boxed{\frac{1}{2\sqrt{2}}\tan^{-1}\frac{t^2}{\sqrt{2}}+C}$