How to pick the right “u” for u substitution and when do I know I need to use it

calculusimproper-integralsintegrationsubstitution

This is the problem I'm working with:

$$\int_{-\infty}^{\infty} \frac{4}{16+x^2}dx$$

So when integrating $4/(16+x^2)$ I know it has to be $\arctan$ something because of the $1/(1+x^2)$ format but how do I exactly know what it is? Like this one guy solved it where he did this:

Let $u=x/4 \Rightarrow 4du=dx$ so the integrand becomes $16/(16+16u^2) = 1/(1+u^2)$ and then it fits right in the $\arctan$ format.

My questions are:

  1. What to pick $u$ when it isn’t as clear? (e.g. How do I come up with the $x/4$?)
  2. Are there any tricks to spot it at first glance?
  3. Are there practice problems to help me gain these skills?
  4. When do I need to use $u$-sub?

Best Answer

Unfortunately for everyone there's no general way of doing this. You have to "see it". For most of us which haven't been integrating since kindergarten, a way to "see it" is to do plenty of exercises. That's how you just "know" $\mathbf{arctan}$ is involved (edit: OP said so in the original version post).

Note that the $4$ in the numerator isn't really an issue, you can take that "outside the integral", informally speaking. So essentially we would like to transform $16+x^2$ into something of the sort of $1+u^2$. One could try to first get rid of that $16$. Again, multiplying or dividing by constants isn't an issue because if we know how to integrate $f$ we know how to integrate $16 \cdot f$.

Well, we might then just deal with $(16+x^2)/16 = 1+x^2/16$. This is not quite right, but then we "know" (as in, we've exposed ourselves to this a billion times already) that $16 = 4^2$, so this is in fact $1+(x/4)^2$. Thus taking $u = x/4$ will help, we just have to deal with the constants we set aside along the way.

I don't know how enlightening this is - and I suspect not very much - but I don't know if there's really much more to this than "guessing".

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