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Chapter 6: Problem 34

Find each integral by using the integral table on the inside back cover. $$ \int x^{2} \sqrt{x^{6}+1} d x $$

Short Answer

Expert verified
The integral is \( \frac{1}{6}x^3\sqrt{x^6+1} + \frac{1}{6}\ln|x^3 + \sqrt{x^6+1}| + C \).

Step by step solution

01

Choose a Suitable Integration Formula

First, we need to identify a formula from the integral table that matches the structure of our integral, \( \int x^{2} \sqrt{x^{6}+1} \, dx \). A common transformation involves substituting a part of the expression to simplify it.
02

Use Substitution Method

Let's use substitution where \( u = x^3 \), which implies \( du = 3x^2 \, dx \) or equivalently \( x^2 \, dx = \frac{1}{3} du \). This substitution simplifies our integral to \( \int \sqrt{u^2 + 1} \frac{1}{3} \, du \).
03

Apply Appropriate Integral Table Formula

Now, look for a formula in the integral table that matches \( \int \sqrt{u^2 + 1} \, du \). The table often includes \( \int \sqrt{a^2 + u^2} \, du \) which has a specific formula to solve, usually involving inverse hyperbolic functions or direct integration steps.
04

Integrate Using the Formula

Using the formula \( \int \sqrt{u^2 + 1} \, du = \frac{1}{2}u\sqrt{u^2+1} + \frac{1}{2}\ln|u + \sqrt{u^2 + 1}| + C \), apply it to our integral. Thus, \( \int \sqrt{u^2 + 1} \frac{1}{3} \, du = \frac{1}{6}u\sqrt{u^2+1} + \frac{1}{6}\ln|u + \sqrt{u^2 + 1}| + C \).
05

Substitute Back to Original Variable

With \( u = x^3 \), substitute back to get the integral in terms of \( x \): \( \frac{1}{6}(x^3)\sqrt{(x^3)^2+1} + \frac{1}{6}\ln|x^3 + \sqrt{(x^3)^2 + 1}| + C \). Simplify to \( \frac{1}{6}x^3\sqrt{x^6+1} + \frac{1}{6}\ln|x^3 + \sqrt{x^6 + 1}| + C \).

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Key Concepts

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

Integration Techniques
In integral calculus, integration techniques help us solve complex integrals. Various methods transform difficult integrals into simpler forms. Some common integration techniques include:
  • Substitution Method
  • Integration by Parts
  • Partial Fraction Decomposition
  • Trigonometric Substitution
The choice of technique depends on the form of the integral. Recognizing the structure within an integral is key.
For instance, if the integral involves compositions like products or quotients of polynomials and roots, substitution is often useful.
This method can simplify the expressions, making integration manageable and better aligned with integral tables.
Substitution Method
The substitution method simplifies an integral by changing variables. It involves finding a substitution that reduces the integral to a standard form.
For example, in this problem, we substitute a part of the expression:
  • Let \( u = x^3 \)
  • This yields \( du = 3x^2 \, dx \) and so \( x^2 \, dx = \frac{1}{3} du \)
The new expression becomes \( \int \sqrt{u^2 + 1} \frac{1}{3} \, du \).
This method transforms our complicated integral into a simpler form. After solving the integral, we can revert to the original variable.
The overall process involves three main steps:
  • Choose a substitution
  • Transform the integral
  • Apply the integral solution and substitute back
Integral Table
An integral table is a valuable tool for solving integrals. It consists of a collection of standard integral formulas used for difficult problems.
When you have transformed an integral into a basic form, you can match it with a corresponding formula in the table. In this exercise:
  • The integral \( \int \sqrt{u^2 + 1} \, du \) is simplified using a formula like \( \int \sqrt{a^2 + u^2} \, du \).
  • Typically, these formulas involve known functions like logarithms, exponentials, or inverse trigonometric functions.
Using the table formula, we integrate the expression to find the solution.
Providing exact results with integral tables prevents computation errors and saves time. Understanding how to use these tables is crucial for anyone studying integral calculus.

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Most popular questions from this chapter

73-74. GENERAL: Permanent Endowments The formula for integrating the exponential function \(a^{b x}\) is \(\int a^{b x} d x=\frac{1}{b \ln a} a^{b x}+C\) for constants \(a>0\) and \(b,\) as may be verified by using the differentiation formulas on page 289. Use the formula above to find the size of the permanent endowment needed to generate an annual \(\$ 12,000\) forever at \(6 \%\) interest compounded annually. [Hint: Find \(\left.\int_{0}^{\infty} 12,000 \cdot 1.06^{-x} d x .\right]\) Compare your answer with that found in Exercise 43 (page 413 ) for the same interest rate but compounded continuously.

\(46-48 .\) BUSINESS: Capital Value of an Asset The capital value of an asset is defined as the present value of all future earnings. For an asset that may last indefinitely (such as real estate or a corporation), the capital value is $$ \left(\begin{array}{c} \text { Capital } \\ \text { value } \end{array}\right)=\int_{0}^{\infty} C(t) e^{-r t} d t $$ where \(C(t)\) is the income per year and \(r\) is the continuous interest rate. Find the capital value of a piece of property that will generate an annual income of \(C(t),\) for the function \(C(t)\) given below, at a continuous interest rate of \(5 \%\). $$ C(t)=59 t^{0.1} \text { thousand dollars } $$

a. Evaluate it by integration by parts. (Give answer in its exact form.) b. Verify your answer to part (a) using a graphing calculator. \(\int_{0}^{2} x^{2} e^{x} d x\)

a. Find the integral \(\int x^{-1} d x\) by integration by parts (using \(u=x^{-1}\) and \(d v=d x\) ), obtaining $$ \int x^{-1} d x=x^{-1} x-\int\left(-x^{-2}\right) x d x $$ which gives $$ \int x^{-1} d x=1+\int x^{-1} d x $$ b. Subtract the integral from both sides of this last equation, obtaining \(0=1 .\) Explain this apparent contradiction.

\(61-64\). Which of the two limits exists? a. \(\lim _{x \rightarrow-\infty} e^{3 x}\) b. \(\lim _{x \rightarrow-\infty} e^{-3 x}\)
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