/*! 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 5 Evaluate the following integrals... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 5

Evaluate the following integrals : $$\int \frac{x^{2}-1}{x^{2}+1} \cdot \frac{d x}{\sqrt{x^{4}+1}}$$

Short Answer

Expert verified
Question: Evaluate the integral: $$\int \frac{x^{2}-1}{x^{2}+1} \cdot \frac{d x}{\sqrt{x^{4}+1}}$$ Answer: The value of the integral is: $$\frac{1}{4} \ln\left|\frac{x^4 + 2x^2 +2}{x^4 -x^2 +2}\right| + C$$

Step by step solution

01

Simplify the expression

Given expression: $$\int \frac{x^{2}-1}{x^{2}+1} \cdot \frac{d x}{\sqrt{x^{4}+1}}$$ Now, let's rewrite the expression in terms of the integrand: $$\int \frac{(x^{2} -1)(dx)}{(x^{2} +1)\sqrt{x^{4}+1}}$$
02

Choose a substitution method

We will use the substitution method to evaluate the integral, but we need to look for a suitable substitution to simplify the expression. Let \(u = x^2\). Then, \(du = 2xdx\) and \(dx = \frac{du}{2x}\). Keeping this in mind, let's substitute \(u\) in the given integral: $$\int \frac{(u -1)\frac{du}{2x}}{(u +1)\sqrt{u^2+1}}$$ Now, we have an integral of a simpler function: $$\frac{1}{2} \int \frac{(u -1) du}{(u +1)\sqrt{u^2+1}}$$
03

Evaluate the integral

To evaluate the integral, let's perform another substitution: Let \(v = u^2 +1\), then \(dv = 2u du\) and \(du = \frac{dv}{2u}\). The integral becomes: $$\frac{1}{2} \int \frac{(u -1)\frac{dv}{2u}}{(u +1)\sqrt{v}}$$ Which simplifies to: $$\frac{1}{4} \int \frac{(u -1) dv}{(u +1)\sqrt{v}}$$ Now, we substitute back \(u\) in terms of \(x\): $$\frac{1}{4} \int \frac{(x^2 -1) dv}{(x^2 +1)\sqrt{v}}$$ By evaluating the integral (notice that it's a standard form), we get: $$\frac{1}{4} \ln\left|\frac{v + (x^2 +1)}{v - (x^2 -1)}\right| + C$$
04

Write the final answer

Now let's substitute back \(v\) in terms of \(x\): $$\frac{1}{4} \ln\left|\frac{x^4 + 2x^2 +2}{x^4 -x^2 +2}\right| + C$$ The final answer is: $$\frac{1}{4} \ln\left|\frac{x^4 + 2x^2 +2}{x^4 -x^2 +2}\right| + C$$

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 91影视!

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

Evaluate the following integrals : $$ \int \frac{x^{5} d x}{\left(1+x^{3}\right)^{1 / 2}} $$

Evaluate \(\int \frac{9 x^{3}-3 x^{2}+2}{\sqrt{3 x^{2}-2 x+1}} d x\)

\(\int \frac{x^{2}-7 x+1}{\sqrt[3]{2 x+1}} d x\)

Evaluate the following integrals: (i) \(\int \frac{d x}{(1+x)^{3 / 2}+(1+x)^{1 / 2}}\) (ii) \(\int \frac{\mathrm{dx}}{\sqrt[4]{5-x}+\sqrt{5-x}}\) (iii) \(\int \frac{\mathrm{dx}}{\sqrt{(\mathrm{x}+2)}+\sqrt[4]{(\mathrm{x}+2)}}\) (iv) \(\int \frac{\sqrt{x+1}+2}{(x+1)^{2}-\sqrt{x+1}} d x\)

Evaluate the following integrals: (i) \(\int \frac{x^{4}}{(1-x)^{3}} d x\) (ii) \(\int \frac{6 x^{2}-12 x+4}{x^{2}(x-2)^{2}} d x\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation
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.