Though it may be basic, computing the antiderivative of ln(x) requires the use of some important, fundamental concepts of calculus. In this Buzzle post, we tell you what these concepts are, and how you can use them to find the antiderivative of ln(x).
Did You Know?
The word 'calculus' is a Latin word, which means a small stone used for counting. Today, calculus refers to any method or system of calculation that comprises symbolic manipulation of the different expressions.
On seeing the apple fall, Newton realized that it was the work of gravity. He also argued that it must be gravity itself that was responsible for the motion of celestial bodies like the comets.
However, when he tried to apply the mathematics of his time to prove his theories, he found it to be inadequate. So, he decided to invent a brand new branch of mathematics―which he called infinitesimal calculus―to support his argument.
To say that Newton was successful in his endeavor is surely an understatement. Calculus not only allowed him to prove his groundbreaking theories of gravity and motion, but also went on to become an inseparable part of modern science.
Today, it finds extensive applications in every branch of physical science, actuarial science, computer science, statistics, engineering, economics, business, medicine, demography etc.
But learning one of the greatest works in mathematics definitely cannot be easy. It is best to take a step-by-step approach, understanding, and applying the different concepts in it, one at a time.
Assuming that you are already familiar with the basics of calculus, including the concepts of derivative and antiderivative (integration), we will show you how to find the antiderivative of ln(x), which will help you learn a few important concepts in calculus. It will prove helpful to you on your journey towards mastering this amazing branch of mathematics.
Integral of Ln(x) by Parts
For finding out the antiderivative of ln(x), you need to first understand the concept of integration by parts. The following is a step-by-step derivation of the formula for integration by parts.
We will start by using the product rule of derivatives. Consider that there are two functions - f(x) and g(x), and we have to find the derivative of the multiplication of those two. This is represented as following.
The equation obtained in Step 3 is the formula for computing integration by parts. It is frequently used for solving difficult integration problems.
To make the integration by parts formula easier to remember, the following substitutions are used.
f(x) = u
g(x)'dx = dv
f'(x)dx = du
g(x) = v
Thus, the formula becomes,
∫ udv = uv - ∫ vdu
To integrate by parts, keep in mind to choose 'u' and 'dv' strategically, so that the calculation becomes easy.
How to Find the Antiderivative of Ln(x)
Now that we are clear with the concept of integration by parts, let us see how we can use it to find the antiderivative that is, the integral of ln(x)dx.
∫ ln(x) dx
To begin with, we first need to choose u and dv. On first inspection, it might appear that there is only one term in the expression. However, since any number multiplied by 1 yields the same number again, we can substitute the following-
∫ ln(x) · 1dx
Now, let us make the following two assignments.
u = ln(x) ----------(a)
dv = 1 · dx -----------(b)
Taking the derivative of u, we get,
du = (1/x) dx ------------(c)
and taking the integration of du, we get,
∫ du = ∫ dv
u = x ------------(d)
We have the formula for integration by parts as,
∫ udv = uv - ∫ vdu
Substituting values from equations (a), (b), (c), and (d) into the formula for integration by parts, we get,
∫ ln(x) dx = ln(x) · x dx - ∫ 1/x · x dx
= x * ln(x) - ∫ dx
= x * ln(x) - x + C
The expression is the antiderivative or the integration of the natural logarithm of x. To verify our answer, let us take its derivative to see if we get back the function ln(x).
[x * ln(x) - x + C]' = ln(x) + (x/x) - 1 + 0
= ln(x) + 1 - 1 + 0
Thus, we can easily find the antiderivative of ln (x) by using integration by parts. This concept is an important one in calculus, which you will frequently be using as you go deeper into it. Hence, we recommend that you solve enough examples to get a firm grip on the important process of integration by parts.