Paul's Online Notes
Home / Calculus II / Integration Techniques / Integrals Involving Roots
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 1-5 : Integrals Involving Roots

In this section we’re going to look at an integration technique that can be useful for some integrals with roots in them. We’ve already seen some integrals with roots in them. Some can be done quickly with a simple Calculus I substitution and some can be done with trig substitutions.

However, not all integrals with roots will allow us to use one of these methods. Let’s look at a couple of examples to see another technique that can be used on occasion to help with these integrals.

Example 1 Evaluate the following integral. $\int{{\frac{{x + 2}}{{\sqrt{{x - 3}}}}\,dx}}$
Show Solution

Sometimes when faced with an integral that contains a root we can use the following substitution to simplify the integral into a form that can be easily worked with.

$u = \sqrt{{x - 3}}$

So, instead of letting $$u$$ be the stuff under the radical as we often did in Calculus I we let $$u$$ be the whole radical. Now, there will be a little more work here since we will also need to know what $$x$$ is so we can substitute in for that in the numerator and so we can compute the differential, $$dx$$. This is easy enough to get however. Just solve the substitution for $$x$$ as follows,

$x = {u^3} + 3\hspace{0.75in}dx = 3{u^2}\,du$

Using this substitution the integral is now,

\begin{align*}\int{{\frac{{\left( {{u^3} + 3} \right) + 2}}{u}\,3{u^2}du}} & = \int{{3{u^4} + 15u\,du}}\\ & = \frac{3}{5}{u^5} + \frac{{15}}{2}{u^2} + c\\ & = \frac{3}{5}{\left( {x - 3} \right)^{\frac{5}{3}}} + \frac{{15}}{2}{\left( {x - 3} \right)^{\frac{2}{3}}} + c\end{align*}

So, sometimes, when an integral contains the root $$\sqrt[n]{{g\left( x \right)}}$$ the substitution,

$u = \sqrt[n]{{g\left( x \right)}}$

can be used to simplify the integral into a form that we can deal with.

Let’s take a look at another example real quick.

Example 2 Evaluate the following integral. $\int{{\frac{2}{{x - 3\sqrt {x + 10} }}\,dx}}$
Show Solution

We’ll do the same thing we did in the previous example. Here’s the substitution and the extra work we’ll need to do to get $$x$$ in terms of $$u$$.

$u = \sqrt {x + 10} \hspace{0.5in}x = {u^2} - 10\hspace{0.5in}dx = 2u\,du$

With this substitution the integral is,

$\int{{\frac{2}{{x - 3\sqrt {x + 10} }}\,dx}} = \int{{\frac{2}{{{u^2} - 10 - 3u}}\left( {2u} \right)\,du}} = \int{{\frac{{4u}}{{{u^2} - 3u - 10}}\,du}}$

This integral can now be done with partial fractions.

$\frac{{4u}}{{\left( {u - 5} \right)\left( {u + 2} \right)}} = \frac{A}{{u - 5}} + \frac{B}{{u + 2}}$

Setting numerators equal gives,

$4u = A\left( {u + 2} \right) + B\left( {u - 5} \right)$

Picking value of $$u$$ gives the coefficients.

\begin{align*}u = & - 2 & \hspace{0.5in} - 8 = & \, B\left( { - 7} \right) & \hspace{0.5in}B = & \, \frac{8}{7}\\ u = & \,5 & \hspace{0.5in} 20 = & \, A\left( 7 \right) & \hspace{0.5in}A = & \, \frac{{20}}{7}\end{align*}

The integral is then,

\begin{align*}\int{{\frac{2}{{x - 3\sqrt {x + 10} }}\,dx}} & = \int{{\frac{{\frac{{20}}{7}}}{{u - 5}} + \frac{{\frac{8}{7}}}{{u + 2}}\,du}}\\ & = \frac{{20}}{7}\ln \left| {u - 5} \right| + \frac{8}{7}\ln \left| {u + 2} \right| + c\\ & = \frac{{20}}{7}\ln \left| {\sqrt {x + 10} - 5} \right| + \frac{8}{7}\ln \left| {\sqrt {x + 10} + 2} \right| + c\end{align*}

So, we’ve seen a nice method to eliminate roots from the integral and put it into a form that we can deal with. Note however, that this won’t always work and sometimes the new integral will be just as difficult to do.