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Chapter 7: Problem 21

Evaluate the following integrals. $$\int \frac{1}{x^{2} \sqrt{x^{2}+9}} d x$$

Short Answer

Expert verified
The value of the integral is \(-(x^2 + 9)^{-\frac{1}{2}} + C\).

Step by step solution

01

Rewrite the integrand

Let's rewrite the given integral, \(\displaystyle\int\frac{1}{x^2\sqrt{x^2+9}}dx\), as follows: $$\int\frac{1}{x^{2} (x^{2} + 9)^{\frac{1}{2}}} d x$$ 2. Choose an appropriate substitution
02

Choose a substitution

We can make a substitution \(u = x^2 + 9\), so \(d u = 2 x d x\). Notice that with this substitution we can easily rewrite the integral in terms of \(u\). 3. Make the substitution and rewrite the integral
03

Make the substitution

Using the substitution \(u = x^2 + 9\), we rewrite the integral in terms of \(u\): $$\frac{1}{2}\int\frac{d u}{u^{\frac{3}{2}}} = \frac{1}{2}\int u^{-\frac{3}{2}} d u$$ 4. Integrate
04

Integrate

Now we integrate the new integrand: $$\frac{1}{2}\int u^{-\frac{3}{2}} d u = \frac{1}{2} \times \frac{u^{-\frac{1}{2}}}{-\frac{1}{2}} = -u^{-\frac{1}{2}}$$ 5. Substitute back to x
05

Substitute back

Now replace \(u\) with the original expression in terms of \(x\): $$ -u^{-\frac{1}{2}} = -(x^2 + 9)^{-\frac{1}{2}}$$ 6. Add the constant of integration
06

Add the constant of integration

Finally, add the constant of integration \(C\): $$-(x^2 + 9)^{-\frac{1}{2}} + C$$ So, the evaluated integral is: $$\int\frac{1}{x^2\sqrt{x^2+9}}dx = -(x^2 + 9)^{-\frac{1}{2}} + C$$

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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 a commonly used technique in integral calculus to simplify expressions before integrating. Think of it as a tool to "change the variable" in a problem to make the integration easier. In this exercise, we deal with the integral \( \int \frac{1}{x^{2} \sqrt{x^{2}+9}} dx \). Here, choosing a smart substitution is the key.

  • The original integral involves a complex expression \( \sqrt{x^{2} + 9} \). By setting \( u = x^2 + 9 \), we handle this complexity more efficiently.
  • After the substitution, the derivative \( du = 2x \, dx \) allows us to express \( dx \) in terms of \( du \), simplifying the integral drastically.
  • This transformation turns our integral into a simpler form: \( \frac{1}{2}\int u^{-\frac{3}{2}} du \), which is straightforward to evaluate.
Such clever substitution reduces the problem to a more soluble form, akin to "breaking down" the problem into smaller, manageable parts.
Definite Integrals
Definite integrals involve boundaries and calculate the net area under the curve of a function. Though, in this exercise, we tackled an indefinite integral, understanding definite integrals is crucial in calculus. To grasp these better:
  • A definite integral \( \int_{a}^{b} f(x) \ dx \) computes the total area under \( f(x) \) as \( x \) moves from \( a \) to \( b \).
  • It outputs a specific numeric value, unlike an indefinite integral, which results in a family of functions \( + C \).
  • The Fundamental Theorem of Calculus links the definite integral of a function to its antiderivative, acting as a bridge between differential and integral calculus.
Although not used directly here, definite integrals underpin many practical applications like finding areas, volumes, and solving physical problems.
Integration Techniques
Integration techniques are strategies to solve integrals efficiently, like substitution or by-parts method. In this example, the substitution method was key. Here we outline common techniques to strengthen your calculus toolkit.

  • Substitution: Simplifies integrals by changing the variable. Crucial for handling nested functions or roots, as seen here.
  • Integration by Parts: Useful where the product of functions is involved. It builds on the product rule of differentiation.
  • Partial Fraction Decomposition: Applied to rational functions, breaking them into simpler fractions for easy integration.
  • Trigonometric Substitution: Effective when dealing with roots of sums or differences of squares.
  • Numerical Integration: Utilized when functions are challenging to integrate analytically, using methods like the trapezoidal or Simpson's rule.
Mastering these techniques equips you for tackling diverse and complex integrals, enhancing problem-solving flexibility and understanding in calculus.

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Most popular questions from this chapter

The work required to launch an object from the surface of Earth to outer space is given by \(W=\int_{R}^{\infty} F(x) d x,\) where \(R=6370 \mathrm{km}\) is the approximate radius of Earth, \(F(x)=G M m / x^{2}\) is the gravitational force between Earth and the object, \(G\) is the gravitational constant, \(M\) is the mass of Earth, \(m\) is the mass of the object, and \(G M=4 \times 10^{14} \mathrm{m}^{3} / \mathrm{s}^{2}.\) a. Find the work required to launch an object in terms of \(m.\) b. What escape velocity \(v_{e}\) is required to give the object a kinetic energy \(\frac{1}{2} m v_{e}^{2}\) equal to \(W ?\) c. The French scientist Laplace anticipated the existence of black holes in the 18th century with the following argument: If a body has an escape velocity that equals or exceeds the speed of light, \(c=300,000 \mathrm{km} / \mathrm{s},\) then light cannot escape the body and it cannot be seen. Show that such a body has a radius \(R \leq 2 G M / c^{2} .\) For Earth to be a black hole, what would its radius need to be?

Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}[\cos ((m-n) x)-\cos ((m+n) x)]\\\&\sin m x \cos n x=\frac{1}{2}[\sin ((m-n) x)+\sin ((m+n) x)]\\\&\cos m x \cos n x=\frac{1}{2}[\cos ((m-n) x)+\cos ((m+n) x)]\end{aligned}$$ $$\int \sin 5 x \sin 7 x d x$$

Recall that the substitution \(x=a \sec \theta\) implies that \(x \geq a\) (in which case \(0 \leq \theta<\pi / 2\) and \(\tan \theta \geq 0\) ) or \(x \leq-a\) (in which case \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\) ). Evaluate for \(\int \frac{\sqrt{x^{2}-1}}{x^{3}} d x,\) for \(x>1\) and for \(x<-1\)

Determine whether the following statements are true and give an explanation or counterexample. a. To evaluate \(\int \frac{4 x^{6}}{x^{4}+3 x^{2}} d x\), the first step is to find the partial fraction decomposition of the integrand. b. The easiest way to evaluate \(\int \frac{6 x+1}{3 x^{2}+x} d x\) is with a partial fraction decomposition of the integrand. c. The rational function \(f(x)=\frac{1}{x^{2}-13 x+42}\) has an irreducible quadratic denominator. d. The rational function \(f(x)=\frac{1}{x^{2}-13 x+43}\) has an irreducible quadratic denominator.

An important function in statistics is the Gaussian (or normal distribution, or bell-shaped curve), \(f(x)=e^{-a x^{2}}.\) a. Graph the Gaussian for \(a=0.5,1,\) and 2. b. Given that \(\int_{-\infty}^{\infty} e^{-a x^{2}} d x=\sqrt{\frac{\pi}{a}},\) compute the area under the curves in part (a). c. Complete the square to evaluate \(\int_{-\infty}^{\infty} e^{-\left(a x^{2}+b x+c\right)} d x,\) where \(a>0, b,\) and \(c\) are real numbers.
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