As a member, you'll also get unlimited access to over 75,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed.
Try it risk-freeAlready registered? Log in here for access
Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.
The integral of ln(x) may look simple, but it's actually a bit involved. To find this integral, we have to use integration by parts. This process is used to find the integral of a product of functions. The formula we use for integration by parts is as follows:
 |
Now you may look at our problem, solve the integral of ln(x), and wonder how this is a product of functions. Well, we can think of the integral of ln(x) as the integral of ln(x)*1. This way, we have a product of the functions f(x) = ln(x) and g(x) = 1.
 |
Okay, now that we've cleared that up, let's look at the steps involved when using integration by parts.
Let's apply these steps to the integral of ln(x). We know our two functions are ln(x) and 1. Since the derivative of ln(x) is well-known as 1/x, it would probably be a good idea to let u = ln(x). Similarly, the integral of 1 is well-known as x + C, where C is a constant. Thus, we will let dv = 1dx. It's important to note that we don't include the constants when finding different integrals during the solving process. This is because the constants that would show up throughout will all be taken care of at the end of the process when we have our final constant.
At this point, we've actually completed steps 1 and 2, and we have our u, du, v, and dv.
| u = ln(x) | dv = 1dx |
| du = (1/x)dx | v = x |
All we have to do now is plug the results into our formula and simplify.
 |
We see that the integral of ln(x) is xln(x) - x + C.
 |
So we've found the integral of ln(x), but the use of integration by parts may be new to you, and it may have left you with some questions. To remedy this, let's take a closer look at integration by parts.
As we said, integration by parts is used to find the integral of products of functions. We can actually derive the formula for integration by parts from the product rule for derivatives. Let's see how this is done.
We'll start with the product formula for derivatives.
 |
Next, we'll integrate both sides of the function. You may wonder why, but this approach will all become more and more clear as we move along.
 |
We can simplify the left hand side of this equation fairly easily. Since finding the integral of something is the same as finding the anti-derivative, we have that the integral of the derivative of f(x)*g(x) is f(x)*g(x) + C. Again, we can drop the constant, because the constants that come up as we go along will be taken care of at the end of the process where they all end up as one constant. Thus, we have the following:
 |
Now, according to the integration by parts formula given earlier, we're finding the integral of u dv. If we let u = f(x) and dv = g ' (x)dx, do you notice any of the integrals in our formula that look like the integral of u dv? Well, looking at the equation, the last integral has the function f times the derivative of the function g. A-ha! This integral would be the integral of u dv where u is f ' (x) and dv is g(x)dx. Therefore, let's solve for this integral using the formula we've got.
 |
All right, we've got our formula. Wait a second, though. This doesn't look like our original integration by way of the parts formula. Don't worry! We can make a few substitutions here, and it will all become crystal clear.
| u = f ' (x) | dv = g ' dx |
| du = f ' (x)dx | v = g(x) |
Plugging these into the formula, we have the following:
 |
There's that formula! Now we see where the integration by parts formula comes from and why we can use it. If all of the details in the derivation of the formula weren't completely clear, that's okay as long as you recognize that it was derived from the product rule for derivatives.
Knowing where a formula comes from can better our understanding of the formula itself and help us identify connections between different mathematical concepts and ideas. Now, not only do you know how to find the integral of ln(x), but you also know why we can use the solving process of integration by parts to do so.
To unlock this lesson you must be a Study.com Member.
Create your account
Already a member? Log In
BackAlready registered? Log in here for access
Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
To learn more, visit our Earning Credit Page
Not sure what college you want to attend yet? Study.com has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.
Back To Course
AP Calculus AB & BC: Tutoring Solution17 chapters | 143 lessons
{{courseNav.course.topics.length}} chapters | {{courseNav.course.mDynamicIntFields.lessonCount}} lessons | {{course.flashcardSetCount}} flashcard set{{course.flashcardSetCoun > 1 ? 's' : ''}}
| 30 day money back guarantee | |||
|
Starting Original Price |
/yr | /mo | |
| Just Just | /day | ||
| Discount For months |
- % - | / | |
|
Price after trial Starting Price starting today |
/ | ||
| Just Just | /day | ||
| Save % off monthly plan | /mo | ||
| Just | /mo | ||
| Price | /yr | ||
Your Cart is Empty. Please Choose a Product.
Study.com video lessons have helped over 30 million students.
"I learned more in 10 minutes than 1 month of chemistry classes"
- Ashlee P.
"I aced the CLEP exam and earned 3 college credits!"
- Clair S.
Over 65 million users have prepared for {{displayNameByProductKey[registrationData.product || cocoon]}} and other exams on Study.com
"The videos have changed the way I teach! The videos on Study.com accomplish in 5 minutes what would take me an entire class."
- Chris F.
Over 65 million users have prepared for {{displayNameByProductKey[registrationData.product || cocoon]}} and other exams on Study.com
Over 65 million users have prepared for {{displayNameByProductKey[registrationData.product || cocoon]}} and other exams on Study.com
