report

How to Integrate ln(sqrt(x)) -- Easy Integral

Students in Calc II (integral calculus) often have trouble finding the antiderivatives of nested functions, also known as function compositions. The function f(x) = ln(sqrt(x)) is an example of a function composition involving the natural logarithm and square root. However, this function is easy to integrate once you realize you can apply the rules of logarithms and exponents to simplify the function.

In contrast, the function g(x) = sqrt(ln(x)), which looks very similar, does not have a closed-form antiderivative. When you are dealing with nested functions, the order in which they are nested is very important.

Step 1: Apply the Identity sqrt(x) = x^(1/2)

The first thing to notice is that taking the square root of a number is the same as raising it to the 1/2 power. Thus, sqrt(x) = x^(1/2). This means that ln(sqrt(x)) is the same as ln(x^(1/2)).

Step 2: Apply the Identity ln(x^a) = a*ln(x)

The next thing is to remember one of the fundamental rules of logarithms: the log of a number raised to a power is the same as that power times the log of the number. In mathematical notation, this is ln(x^(1/2)) = (1/2)*ln(x).

Step 3: Integrate by Parts

Since we have seen that ln(sqrt(x)) is equivalent to (1/2)*ln(x), the process of finding the antiderivative is straight forward using the technique of integration by parts. Integration by parts is discussed in more detail another tutorial. We will work it out here in less detail assuming you already know the method.

Let

u = ln(x), and
dv = dx,

which gives you

du = 1/x dx, and
v = x.

The integral of (1/2)*ln(x) is then

(1/2) * ∫ ln(x) dx
= (1/2) * ∫ u dv
= (1/2) * [uv - ∫ v du]
= (1/2) * [x*ln(x) - ∫ dx]
= (1/2) * [x*ln(x) - x]

In summary, the exact integral of ln(sqrt(x)) is

∫ ln(sqrt(x)) dx = (1/2) * [x*ln(x) - x] + C

where C is an arbitrary constant.

Example: Find the Exact Area Under a Curve

Graph of y = 3*ln(sqrt(x)) with area under curve shaded between x = 4e and x = 5e.
Graph of y = 3*ln(sqrt(x)) with area under curve shaded between x = 4e and x = 5e.

Using the integral of ln(sqrt(x)), we can find the exact area under the curve y = 3*ln(sqrt(x)) from x = 4e to x = 5e. (Recall, e is the natural logarithm base 2.7182818...) The antiderivative of 3*ln(sqrt(x)) works out to be

∫ 3*ln(sqrt(x)) dx
= (3/2)*[x*ln(x) - x]

Plugging in the limits x = 5e and x = 4e and subtracting gives us the exact answer

(3/2)[5e*ln(5e) - 5e] - (3/2)[4e*ln(4e) - 4e]
= (3e/2)ln(3125/256)

This is approximately 10.20176 rounded to five decimal places.

Another Example

Problem: The velocity of a particle as a function of time is v(t) = 4*log(sqrt(2t+1)) m/s, where "log" is the logarithm in base 10 and "m/s" is the velocity unit meters per second. How far does the particle travel between t = 1 second and t = 2 seconds?

Solution: The velocity function involves a base-10 logarithm, but the integral we worked out above involves the natural logarithm, aka logarithm base-e. Therefore, the first step is to use the change of base formula for logarithms:

log_m (x) = ln(x) / ln(m)

The velocity function can be equivalently written as

v(t) = 4*ln(sqrt(2t+1)) / ln(10)

Using the identity ln(x^a) = a*ln(x), we can further simplify the velocity function to

v(t) = [2/ln(10)] * ln(2t+1)

In physics, when you know the velocity of a particle as a function of time, you can find the total distance traveled in an interval by integrating the velocity function with respect to time over that interval. Therefore the last step to solve this problem is to integrate v(t) = [2/ln(10)] * ln(2t+1) from t = 1 to t = 2.

To make the integral simpler, we can use a linear transformation 2t + 1 = u. This gives us dt = 1/2 du, and changes the interval from [1, 2] to [3, 5].

∫ [2/ln(10)] * ln(2t + 1) dt, {1 ≤ t ≤ 2}
= ∫ [2/ln(10)] * ln(u) * (1/2) du, {3 ≤ u ≤ 5}
= ∫ ln(u) / ln(10) du, { 3 ≤ u ≤ 5}
≈ 1.1948972937146

The particle travels about 1.195 meters between t = 1 second and t = 2 seconds.

Related Integral: Find the Antiderivative of ln(sqrt(x) + k)

A problem that is superficially similar to the topic of this tutorial is to find the antiderivative of the function ln(sqrt(x) + k), where k is some constant. Unfortunately this integral calculus problem will not yield to basic logarithm identities; there is no logarithmic identity that will simplify the logarithm of a sum, log(A + B). To work out this integral you need to make the change of variables x = w^2 and dx = 2w dw. This transforms the original integral into

∫ ln(sqrt(x) + k) dx
= ∫ 2w*ln(w + k) dw

With a further change of variable w = v - k and dw = dv, you get

∫ 2w*ln(w + k) dw
= ∫ 2(v - k)*ln(v) dv
= 2*∫ v*ln(v) - k*∫ ln(v) dv

Working this out with integration by parts and making the reverse substitutions gives you

∫ ln(sqrt(x) + k) dx
= (x - k^2)*ln(sqrt(x) + k) + k*sqrt(x) - x/2 + C

As you can see, merely adding a constant inside the logarithm makes integration more difficult.

Non-Integrable Functions with Logs and Roots

The following functions have closed-form antiderivatives, i.e., you can express their antiderivatives in terms of elementary functions.

  • ln(sqrt(x))
  • ln(sqrt(x) + k)
  • sqrt(x)*ln(x)
  • ln(x)/sqrt(x)

Unfortunately, the majority functions involving both roots and logarithms cannot be integrated so cleanly. Here are some functions that don't have closed-form expressions.

  • sqrt(ln(x))
  • sqrt[k + ln(x)]
  • sqrt(x)/ln(x)
  • sqrt(x)/[k + ln(x)]
  • ln(x)/[k + sqrt(x)]
  • e^sqrt(ln(x))

The integrals of these functions can be expressed in terms of non-elementary functions and/or power series. Definite integrals can be computed with numerical algorithms.

More by this Author


Comments

No comments yet.

    Sign in or sign up and post using a HubPages Network account.

    0 of 8192 characters used
    Post Comment

    No HTML is allowed in comments, but URLs will be hyperlinked. Comments are not for promoting your articles or other sites.


    calculus-geometry profile image

    TR Smith (calculus-geometry)213 Followers
    449 Articles

    TR Smith is a mathematician, programmer and product designer who has taught math for 30 years, and uses it every day.



    Click to Rate This Article

    Menu

    Explore