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?

  • $\begingroup$ Do you know what the derivative of $\arctan(u)$ function is and how it differs, if at all, from the derivative of $\ln(1+u^2)$? $\endgroup$ – Dilip Sarwate Jun 5 '14 at 1:12
  • $\begingroup$ I was taught the derivative of arctan at some point but I haven't needed it so I forgot that it is $\frac{1}{x^2+1}$. I tend to forget math very easily. I did a bunch of questions recently with a similar looking integrand and in those cases the integrand was (possibly a constant other than 1)*ln |"denominator of original integrand"| so I assumed that this was the same case. Now that I look at those, they were partial fractions and the denominator was always linear. I suppose the fact that it isn't here is part of the problem. $\endgroup$ – Greener Jun 5 '14 at 5:18

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.

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.

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  • $\begingroup$ How did you know that $u^2$ has to be divided by $\sqrt2$? I would've just factored out a 2 from the whole denominator, get $2(u^2/2 + 1)$ and then left it there. $\endgroup$ – Greener Jun 5 '14 at 6:09
  • $\begingroup$ @Greener Because $u^2/2 + 1$ isn't of the right form yet; the arctangent integral that I wrote down would need to have $(\text{something})^2 + 1$. $\endgroup$ – user61527 Jun 5 '14 at 15:59

Note that:

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

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

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  • $\begingroup$ oh (づ。◕‿‿◕。)づ +1 $\endgroup$ – Lucas Zanella Jun 5 '14 at 4:26

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