Calculus u-substitution, how to know when to use square root term for u?

Question

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!

Answer 1

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$$x=\sqrt3u\iff u=\frac{\sqrt3}3x.$$

Answer 2

The derivative of the function $\arctan(x)$ is given by \begin{align*} \frac{\mathrm{d}\arctan(x)}{\mathrm{d}x} = \frac{1}{1 + x^{2}} \end{align*}

So we are looking for a change of variable which makes the integrand resembles such derivative.

In order to factor out the number $3$, we make the substitution $x = \sqrt{3}u$, whence we get the integral: \begin{align*} \int\frac{\mathrm{d}x}{x^{2} + 3} = \frac{1}{\sqrt{3}}\int\frac{\mathrm{d}u}{1 + u^{2}} = \frac{\arctan(u)}{\sqrt{3}} + c = \frac{1}{\sqrt{3}}\arctan\left(\frac{x}{\sqrt{3}}\right) + c \end{align*}

Hopefully this helps !