Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In the following exercises, find each indefinite integral by using appropriate substitutions. lnxxdx

Short Answer

Expert verified
The indefinite integral is (lnx)22+C.

Step by step solution

01

Recognize the Need for Substitution

Identify the function and the part of it that can be substituted to simplify the integral. In this case, notice that the presence of lnx in the numerator and x in the denominator suggests a substitution method can be applied.
02

Perform the Substitution

Choose u=lnx. Then, compute the derivative of u, which gives du=1xdx. This implies that dx=xdu. Substitute these into the integral, giving udu.
03

Integrate the Substituted Function

Now, integrate udu. Using the power rule for integration, this integral becomes u22+C, where C is the constant of integration.
04

Substitute Back to Original Variable

Replace u back with lnx to return to the original variable's terms. Thus, u22+C becomes (lnx)22+C.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is a powerful tool in calculus, especially for solving integrals. It simplifies complex integrals by transforming them into more manageable forms. In our example, we need to find the indefinite integral of lnxxdx. This setup is perfect for using substitution since the derivative of lnx is 1x, conveniently matching the denominator.

To apply the substitution method, we start by choosing a variable to substitute. We set u=lnx and calculate its differential, which is du=1xdx. This immense simplification allows us to change the variable in the integral from x to u, thus transforming lnxxdx into udu.

This new integral is significantly easier to handle. After integrating, remember to substitute the original variable back into the result. This completes the substitution method, turning complex integrals into simpler ones by temporarily switching variables.
Integration Techniques
Integration techniques are essential tools for solving different types of integrals. They allow us to handle integrals that might not be straightforward at first glance. In this context, using substitution is one such technique. It is especially useful when the integral has a structure that matches a known derivative.

In our exercise, after substituting u=lnx, we integrate udu. This requires the application of a basic integration rule: the power rule. The power rule states that undu=un+1n+1+C for any real number neq1. Here, n=1, so we get udu=u22+C.

This technique simplifies the work, allowing us to solve the integral without directly dealing with the original complex functions. It's an elegant way to break down daunting problems into manageable parts using standard integration rules.
Logarithmic Functions
Logarithmic functions, like lnx, have unique properties that make them interesting and occasionally challenging in integration. They often appear in integrals where substitution is beneficial. Understanding the behavior of logarithmic functions, especially their derivatives, is crucial.

In the integral lnxxdx, the expression involves the natural logarithm lnx. One key property of logarithmic functions is that their derivative, ddx(lnx), is 1x. This property is what makes substitution possible here and transforms the integral into a simpler form.

When dealing with logarithmic functions in integrals, always check if their derivatives might help simplify the problem. This knowledge not only helps reduce complicated calculus challenges but also deepens your understanding of the behavior and characteristics of logarithmic functions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free