Integrals Involving Roots

A root substitution is simply a substitution where you let $u$ be the entire root in the integral, like $u = \sqrt{x + 1}$, instead of something underneath it. It's good for integrals where a root like $\sqrt{x}$ or $\sqrt[3]{x}$ is the only thing making them hard, which happens a lot with square-root laws in physics, like a tank draining. All we're really saying is: give the whole root a new name and the ugliness goes away. The setup $u = \sqrt{x + 1}$, $x = u^2 - 1$, $dx = 2u\,du$ looks like three formulas to memorize, but it's just three small steps: $u = \sqrt{x + 1}$ (name the root), $x = u^2 - 1$ (square both sides and solve, so $x$ is written in $u$) and $dx = 2u\,du$ (differentiate that, so $dx$ is written in $u$ too). After that, every piece of the integral is in terms of $u$, and there's no root anywhere. It's simpler than it looks; bear with it.

Take $\int x\sqrt{x + 1}\,dx$. In the substitution lesson, you'd look for the derivative of the inside, $x + 1$, sitting in the integrand as a factor. Its derivative is $1$, so in a sense it's there, and $w = x + 1$ even works here. But try $\int \frac{dx}{1 + \sqrt{x}}$: there's nothing to cancel, and ordinary substitution gets stuck.

Root substitution doesn't need the derivative to be lying around. Instead of hunting for $du$, you solve for $x$ in terms of $u$ and replace everything, including $dx$. Notice what that does to the root: $\sqrt{x + 1}$ becomes $u$, a plain variable, and $x$ itself becomes the polynomial $u^2 - 1$. Polynomials in $u$ are about the friendliest thing there is to integrate. At worst you get a rational function in $u$, and the previous lesson handles those.

1. Let $u = \sqrt[n]{g(x)}$ (the whole root).
2. Solve for $x$: $x =$ something in $u$.
3. Compute $dx$ in terms of $du$.
4. Rewrite the integrand entirely in $u$. It's usually a polynomial or a rational function now.
5. Integrate and substitute back.

For definite integrals, convert the limits: when $x$ runs from $a$ to $b$, $u$ runs from $\sqrt[n]{g(a)}$ to $\sqrt[n]{g(b)}$.

If several roots appear, like $\sqrt{x}$ and $\sqrt[3]{x}$, let $u = \sqrt[6]{x}$. Six is the smallest number both 2 and 3 divide into, so both roots become whole powers of $u$: $\sqrt{x} = u^3$ and $\sqrt[3]{x} = u^2$.

A common mistake. It's tempting to swap the root for $u$ and leave $dx$ alone, ending up with something like $\int (u^2 - 1)\,u\,dx$. That mixes two variables and gives the wrong answer. The $dx$ has to become $2u\,du$ too, and that extra factor of $u$ matters, as the first worked example below shows.

Roots appear wherever a square-root law does: the speed of an object that falls a height $h$ is $\sqrt{2gh}$, a pendulum's period grows like $\sqrt{L}$, and the time to drain a tank through a hole is an integral of $1/\sqrt{h}$ (Torricelli's law).