How to Integrate Square Root Functions

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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

Integration of Basic Square Root Functions

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}

Integration of More Complex Square Root Functions

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}