Integral of $\frac{t}{t^4+2} dt$

Question

The answer in the back of the calculus book is $$\frac{1}{2\sqrt2}\arctan \left( \frac{t^2}{\sqrt2} \right) + C$$ and I have no idea how they reached this answer. My first guess was to try partial fractions but I don't think I can in this case. I then tried u substitution using $u=t^2$ and $du=2t$, giving me

$$\frac{1}{2}\int \frac{du}{u^2+2}$$

I thought that I'd be able integrate this to reach an answer like $\frac{1}{2} \ln |t^2 + 1| + C$ but that's of course not the case and I'm not sure why. How should I approach this?

Answer 1

For starters, the derivative of $\ln |t^2 + 1|$ is

$$\frac{2t}{t^2 + 1}$$

so there's a problem - that $t$ in the numerator.

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Rather, what you should use is that

$$\int\frac{1}{x^2 + 1} dx = \arctan x + C$$

You can use this form by dividing by $2$ in the denominator, and writing

$$\frac{1}{u^2 + 2} = \frac{1}{2((u/\sqrt 2)^2 + 1)} = \frac 1 2 \frac{1}{(u / \sqrt 2)^2 + 1}$$

Now a simple substitution allows the integral to be evaluated.

Answer 2

Note that:

$$\int\frac{1}{u^2 + 2}du = \frac{1}{2}\int \frac{du}{\left(\frac{u}{\sqrt{2}}\right)^2 + 1} = \frac{1}{\sqrt{2}}\arctan\left(\frac{u}{\sqrt{2}}\right) + C$$

Answer 3

$$ \int \frac{t}{t^4+2}\,dt = \frac 1 {2\sqrt{2}} \int \frac{1}{\left(\underbrace{{}\quad\dfrac{t^2}{\sqrt{2}}\quad{}}_{\large u}\right)^2+1} \left(\underbrace{\sqrt{2}\,t\,dt}_{\large du}\right) $$