How To Integrate Square Root Functions

Integrating functions is one of the core applications of calculus. Sometimes, this is straightforward, as in:

\(F(x) = \int( x^3 + 8) dx\)

In a comparatively complicated example of this type, you can use a version of the basic formula for integrating indefinite integrals:

\(\int (x^n + A) dx = \frac{x^{(n + 1)}}{n + 1} + Ax + C\)

where ​A​ and ​C​ are constants.

Thus for this example,

\(\int x^3 + 8 = \frac{x^4}{4} + 8x + C\)

On the surface, integrating a square root function is awkward. For example, you may be stymied by:

\(F(x) = \int \sqrt{(x^3) + 2x – 7}dx\)

But you can express a square root as an exponent, 1/2:

\(\sqrt{x^3} = x^{3(1/2)} = x^{(3/2)}\)

The integral therefore becomes:

\(\int (x^{3/2} + 2x – 7)dx\)

to which you can apply the usual formula from above:

\(\begin{aligned}
\int (x^{3/2} + 2x – 7)dx &= \frac{x^{(5/2)}}{5/2} + 2\bigg(\frac{x^2}{2}\bigg) – 7x\)
\(&= \frac{2}{5}x^{(5/2)} + x^2 – 7x
\end{aligned}\)

Sometimes, you may have more than one term under the radical sign, as in this example:

\(F(x) = \int \frac{x + 1}{\sqrt{x – 3}}dx\)

You can use ​u​-substitution to proceed. Here, you set ​u​ equal to the quantity in the denominator:

\(u = \sqrt{x – 3}\)

Solve this for ​x​ by squaring both sides and subtracting:

\(u^2 = x – 3\)
\(x = u^2 + 3\)

This allows you to get dx in terms of ​u​ by taking the derivative of ​x​:

\(dx = (2u)du\)

Substituting back into the original integral gives

\(\begin{aligned}\)
\(F(x) &= \int \frac{u^2 + 3 + 1}{u}(2u)du\)
\(&= \int \frac{2u^3 + 6u + 2u}{u}du\)
\(&= \int (2u^2 + 8)du\)
\(\end{aligned}\)

Now you can integrate this using the basic formula and expressing ​u​ in terms of ​x​:

\(\begin{aligned}\)
\(\int (2u^2 + 8)du &= \frac{2}{3}u^3 + 8u + C\)
\(&= \frac{2}{3} (\sqrt{x – 3})^3 + 8( \sqrt{x – 3}) + C\)
\(&= \frac{2}{3} (x – 3)^{(3/2)} + 8(x – 3)^{(1/2)} + C\)
\(\end{aligned}\)