/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Find the integral. $$ \int \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the integral. $$ \int \frac{1}{x \sqrt{x^{4}-4}} d x $$

Short Answer

Expert verified
The evaluated integral is \(\frac{1}{2}\ln|x^{2}/\sqrt{x^{4}-4}+ \sqrt{x^{4}/4-1}|\) + C.

Step by step solution

01

Perform a substitution

Let's perform a substitution to simplify this integral. Set \(u = x^2\). This means \(du = 2x \, dx\), and in terms of \(x \, dx\) is \(du/2\). Thus, the original integral now becomes \(\int \frac{du}{2u \sqrt{u^{2}-4}}\).
02

Further simplify the integral

We can simplify the integral by taking out the constant term from the integral and applying the rule \( \int f(x) \, dx = F(x) + C \), which states that the integral of a function is its antiderivative plus a constant. This gives us: \(\frac{1}{2}\int \frac{du}{u \sqrt{u^{2}-4}}\).
03

Use a standard integral formula

This integral is now in the form of a standard integral: \(\int \frac{du}{u \sqrt{u^{2}-a^{2}}}\) where \(a = 2\). The standard integral has the result: \(\frac{1}{a}\ln|u/\sqrt{u^{2}-a^{2}}+ \sqrt{u^{2}/a^{2}-1}|+C\). Therefore, substituting \(a = 2\), the result is \(\frac{1}{2}\ln|u/\sqrt{u^{2}-4}+ \sqrt{u^{2}/4-1}|\).
04

Substitute back

Substitute \(u = x^2\) back, the result is then \(\frac{1}{2}\ln|x^{2}/\sqrt{x^{4}-4}+ \sqrt{x^{4}/4-1}|\).

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 91Ó°ÊÓ!

One App. One Place for Learning.

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

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.