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Chapter 7: Problem 43

Evaluate the following integrals. $$\int \frac{x^{3}}{\left(81-x^{2}\right)^{2}} d x$$

Short Answer

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Question: Evaluate the given integral: $$\int \frac{x^{3}}{\left(81-x^{2}\right)^{2}} dx$$ Answer: The evaluated integral is: $$-\frac{1}{2}\ln{(81 - x^2)} + C$$

Step by step solution

01

Identify the substitution

Let \(u = 81 - x^2\). Then, \(-2x\, dx = du\).
02

Rewrite the integral in terms of \(u\)

Substituting \(u\) and solving for \(x^3\, dx\), we get: $$x^3\, dx = -\frac{1}{2}u\, du$$ Since \(x^2 = 81 - u\), we can write \(x^3 = x^2 \cdot x = (81 - u)x\). Now, we can substitute this back into the integral: $$\int \frac{1}{u^2}\left(-\frac{1}{2}u\right)\, du = -\frac{1}{2}\int \frac{1}{u}\, du$$
03

Integrate the function with respect to \(u\)

Now, we will integrate the simplified function with respect to \(u\): $$-\frac{1}{2}\int \frac{1}{u}\, du = -\frac{1}{2}\ln{|u|} + C$$
04

Substitute back in terms of \(x\)

Now, we will replace \(u\) with the original substitution \(u = 81 - x^2\) and simplify the expression: $$-\frac{1}{2}\ln{(81 - x^2)} + C$$
05

Write the final answer

The final expression for the evaluated integral is: $$\int \frac{x^{3}}{\left(81-x^{2}\right)^{2}} dx = -\frac{1}{2}\ln{(81 - x^2)} + C$$

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Key Concepts

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

Substitution Method
The substitution method is an essential integration technique used to simplify complex integrals. It involves changing the variable of integration, which converts the integral into a more familiar or easier form to evaluate.

Here's how the process works:
  • Identify a portion of the integral that can be replaced by a single variable, often denoted as \( u \).
  • Determine the differential of this new variable, typically expressed as \( du \). In our example, \( u = 81 - x^2 \) and \(-2x \, dx = du \).
  • Substitute and transform the original integral into terms of \( u \) and \( du \), simplifying the equation.
Applying substitution often makes the integral easier to solve. Once the integration is complete, reverse the substitution to return to the original variable, ensuring that the solution aligns with the original problem. Substitution is particularly useful for handling integrals involving composite functions or products of functions.
Definite Integrals
A definite integral is a way to calculate the area under the curve of a function between two points. This occurs by integrating the function over an interval \([a, b]\). Unlike indefinite integrals, definite integrals result in a specific numerical value.

Characteristics of definite integrals include:
  • Definite integrals use limits, from \(a\) to \(b\), to specify the start and end of the interval of integration.
  • The result gives the net area, considering areas above the x-axis as positive and below as negative.
  • To solve, first find the antiderivative of the function and then evaluate it at the upper and lower limits, \(b\) and \(a\).
To remember, the notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), and it is solved by the formula: \( F(b) - F(a) \), where \( F \) is the antiderivative of \( f(x) \).
Indefinite Integrals
Indefinite integrals refer to finding the antiderivative of a function without specific limits of integration. The result represents a family of functions rather than a single value.

Key points about indefinite integrals are:
  • Unlike definite integrals, they do not have upper or lower bounds.
  • The solution includes a constant \( C \), since integration is the reverse of differentiation and constants disappear during differentiation.
  • The general form is \( \int f(x) \, dx = F(x) + C \), where \( F(x) \) is the antiderivative of \( f(x) \).
Indefinite integrals are highly useful in problems where the exact places for evaluation aren't specified, or the context is focusing more on function shapes and trends rather than specific numerical solutions. They're ubiquitous in solving differential equations and understanding the behavior of functions.

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. The Trapezoid Rule is exact when used to approximate the definite integral of a linear function. b. If the number of subintervals used in the Midpoint Rule is increased by a factor of \(3,\) the error is expected to decrease by a factor of 8. c. If the number of subintervals used in the Trapezoid Rule is increased by a factor of \(4,\) the error is expected to decrease by a factor of 16.

Use symmetry to evaluate the following integrals. a. \(\int_{-\infty}^{\infty} e^{-|x|} d x\) b. \(\int_{-\infty}^{\infty} \frac{x^{3}}{1+x^{8}} d x\)

Arc length of a parabola Let \(L(c)\) be the length of the parabola \(f(x)=x^{2}\) from \(x=0\) to \(x=c,\) where \(c \geq 0\) is a constant. a. Find an expression for \(L\) and graph the function. b. Is \(L\) concave up or concave down on \([0, \infty) ?\) c. Show that as \(c\) becomes large and positive, the arc length function increases as \(c^{2} ;\) that is, \(L(c) \approx k c^{2},\) where \(k\) is a constant.

Graph the integrands and then evaluate and compare the values of \(\int_{0}^{\infty} x e^{-x^{2}} d x\) and \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x\).

Solve the following problems using the method of your choice. $$\frac{d p}{d t}=\frac{p+1}{t^{2}}, p(1)=3$$
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