I am trying to solve $$\int {1\over x^2+3}dx$$
When I look at solutions created by several online solvers, they use u-substitution and choose: $$u={\sqrt{3}\over{3}}x$$
then proceed with $$\int {1\over x^2+3}dx = \int {\sqrt{3}\over 3(u^2+1)}du$$
I've seen this approach used on several other types of problems where $u$ involves using a square root to get the denominator into a form that contains $u^2+1$ (or more generally $u^2+a^2$), but I don't understand how $u$ is being chosen.
I presume this step is more elementary algebra than calculus, but I've clearly missed the material that shows when and how to develop $u$ terms like this.
Would you please explain the steps involved in order to realize $u$ needs to contain a square root term, how to chose the right $u$, and how to apply it to get $u^2+1$ in the denominator?
Thank you very much!
Best Answer
Since we know an antiderivative of$$\frac1{x^2+1}\tag1$$($\arctan(x)$), the idea is to transform $\frac1{x^2+3}$ in such a way that we get something like $(1)$. The simplest way is to do $x=\sqrt3\,u$ and $\mathrm dx=\sqrt3\,\mathrm du$; then$$\int\frac1{x^2+3}\,\mathrm dx$$becomes$$\int\frac{\sqrt3}{3(u^2+1)}\,\mathrm du.$$And$$x=\sqrt3u\iff u=\frac{\sqrt3}3x.$$