5
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Do you know what the derivative of $\arctan(u)$ function is and how it differs, if at all, from the derivative of $\ln(1+u^2)$? $\endgroup$ 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
6
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
3
$\begingroup$

Note that:

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

$\endgroup$
0
2
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.