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Chapter 1 Solutions

Worked solutions to all thirty problems, with the reason for each step. These explanations were added for this edition; the chapter answer key and original worked examples follow below.

Chapter problems →

Problem 1

Original problem · added worked solution

1. Original source equation or figure

The real integrand is defined on $x > 1$ or $x < - 1$ . Let $u = \sqrt{x^{2} - 1}$ , so $d u = \frac{x}{\sqrt{x^{2} - 1}} d x$ and $x^{2} = 1 + u^{2}$ .
The integral becomes $\int \frac{2}{(1 + u^{2})^{2}} d u$ . Put $u = \tan θ$ ; then $d u = \sec^{2} θ d θ$ , giving $\int 2 \cos^{2} θ d θ$ .
Use $2 \cos^{2} θ = 1 + \cos (2 θ)$ and integrate: $θ + \sin θ \cos θ + C$ .
Since $θ = \arctan u$ and $\sin θ \cos θ = \frac{u}{1 + u^{2}}$ , substitute back. The same formula differentiates correctly on each real domain interval.

Final answer

$\arctan (\sqrt{x^{2} - 1}) + \frac{\sqrt{x^{2} - 1}}{x^{2}} + C$ , for $| x | > 1$ .

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Problem 2

Original problem · added worked solution

2. Original source equation or figure

Rewrite $\sqrt{(\frac{x}{5})^{2} + 3} = \frac{\sqrt{x^{2} + 75}}{5}$ , so the integral is $5 \int \frac{1}{\sqrt{x^{2} + 75}} d x$ .
Let $x = 5 \sqrt{3} \tan θ$ with $- \frac{π}{2} < θ < \frac{π}{2}$ . Then the radical is $5 \sqrt{3} \sec θ$ and the integral is $5 \int \sec θ d θ$ .
Integrate to obtain $5 \ln | \sec θ + \tan θ | + C$ . Replacing the trigonometric functions and absorbing the positive constant inside the logarithm into $C$ gives the result.

Final answer

$5 \ln (x + \sqrt{x^{2} + 75}) + C$ , for all real $x$ .

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Problem 3

Original problem · added worked solution

3. Original source equation or figure

Let $I_{m} = \int x^{m} e^{x} d x$ . Integration by parts with $u = x^{m}$ and $d v = e^{x} d x$ gives $I_{m} = x^{m} e^{x} - m I_{m - 1}$ .
Apply this recurrence five times, reducing the polynomial degree at each step: $I_{5} = e^{x} (x^{5} - 5 x^{4} + 20 x^{3} - 60 x^{2} + 120 x) - 120 I_{0}$ .
Since $I_{0} = e^{x}$ , collect the terms. Differentiating $e^{x} P (x)$ gives $e^{x} (P + P')$ ; here the lower degree terms cancel, leaving $x^{5} e^{x}$ .

Final answer

$e^{x} (x^{5} - 5 x^{4} + 20 x^{3} - 60 x^{2} + 120 x - 120) + C$ .

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Problem 4

Original problem · added worked solution

4. Original source equation or figure

Let $I_{m} = \int x^{m} \sin x d x$ . Two integrations by parts give $I_{m} = - x^{m} \cos x + m x^{m - 1} \sin x - m (m - 1) I_{m - 2}$ .
Use the recurrence first at $m = 5$ and then at $m = 3$ : $I_{5} = - x^{5} \cos x + 5 x^{4} \sin x + 20 x^{3} \cos x - 60 x^{2} \sin x + 120 I_{1}$ .
One final integration by parts gives $I_{1} = - x \cos x + \sin x$ . Substitute this expression and include the integration constant.

Final answer

$- x^{5} \cos x + 5 x^{4} \sin x + 20 x^{3} \cos x - 60 x^{2} \sin x - 120 x \cos x + 120 \sin x + C$ .

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Problem 5

Original problem · added worked solution

5. Original source equation or figure

For $x \neq 0, - 5$ , use one term for every power of the repeated linear factor and a linear numerator for the irreducible quadratic: $\frac{1}{x^{5} (x + 5) (x^{2} + 2)} = \frac{A_{1}}{x} + \frac{A_{2}}{x^{2}} + \frac{A_{3}}{x^{3}} + \frac{A_{4}}{x^{4}} + \frac{A_{5}}{x^{5}} + \frac{B}{x + 5} + \frac{D x + E}{x^{2} + 2}$ .
Multiply by $x^{5} (x + 5) (x^{2} + 2)$ and equate polynomial coefficients. This gives $A_{1} = \frac{579}{25000}$ , $A_{2} = \frac{23}{2500}$ , $A_{3} = - \frac{23}{500}$ , $A_{4} = - \frac{1}{50}$ , $A_{5} = \frac{1}{10}$ , $B = - \frac{1}{84375}$ , $D = - \frac{5}{216}$ , and $E = - \frac{1}{108}$ .
Integrate the powers using $\int x^{- k} d x = \frac{x^{1 - k}}{1 - k}$ for $k \neq 1$ , and integrate the two linear-factor reciprocals with logarithms.
Split the quadratic numerator. Use $\int \frac{x}{x^{2} + 2} d x = \frac{1}{2} \ln (x^{2} + 2)$ and $\int \frac{1}{x^{2} + 2} d x = \frac{1}{\sqrt{2}} \arctan (\frac{x}{\sqrt{2}})$ . Combine all contributions.

Final answer

$\frac{579}{25000} \ln | x | - \frac{23}{2500 x} + \frac{23}{1000 x^{2}} + \frac{1}{150 x^{3}} - \frac{1}{40 x^{4}} - \frac{1}{84375} \ln | x + 5 | - \frac{5}{432} \ln (x^{2} + 2) - \frac{1}{108 \sqrt{2}} \arctan (\frac{x}{\sqrt{2}}) + C$ , on intervals avoiding $0$ and $- 5$ .

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Problem 6

Original problem · added worked solution

6. Original source equation or figure

The numerator has higher degree than the denominator. Divide by $x^{3} + 2 x^{2} - x - 2$ to obtain $\frac{x^{5} - 3}{(x^{2} - 1) (x + 2)} = x^{2} - 2 x + 5 + \frac{- 10 x^{2} + x + 7}{(x - 1) (x + 1) (x + 2)}$ .
Write the proper fraction as $\frac{A}{x - 1} + \frac{B}{x + 1} + \frac{D}{x + 2}$ . Multiplying by the denominator and substituting $x = 1, - 1, - 2$ gives $A = - \frac{1}{3}$ , $B = 2$ , and $D = - \frac{35}{3}$ .
Integrate the polynomial term by term and each reciprocal linear factor with a logarithm. The antiderivative is valid on each interval that avoids the three poles.

Final answer

$\frac{x^{3}}{3} - x^{2} + 5 x - \frac{1}{3} \ln | x - 1 | + 2 \ln | x + 1 | - \frac{35}{3} \ln | x + 2 | + C$ , for $x \neq - 2, - 1, 1$ .

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Problem 7

Original problem · added worked solution

7. Find the arclength of Original source equation or figure from .

Differentiate the coordinates: $x' (t) = \cos t$ and $y' (t) = - \sin t$ .
The speed is $\sqrt{\cos^{2} t + \sin^{2} t} = 1$ , by the Pythagorean identity.
Integrate the speed over the displayed interval $- π \leq t \leq π$ : $L = \int_{- π}^{π} 1 d t = 2 π$ .

Final answer

$2 π$ units.

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Problem 8

Original problem · clarified · added worked solution

8. Parametrize an ellipse centered at the origin, with a horizontal major axis of 4 units and a minor axis of 2 units, which has a clockwise orientation

Clarification: the ellipse is centered at the origin and its major axis is horizontal. These conventions make the problem agree with the stated parametrization.

The specified full axis lengths are $4$ and $2$ , so the horizontal and vertical semiaxes are $2$ and $1$ . With center $(0, 0)$ , the ellipse equation is $\frac{x^{2}}{4} + y^{2} = 1$ .
Choose $x (t) = 2 \cos t$ and $y (t) = - \sin t$ , with $0 \leq t \leq 2 π$ . The identity $\cos^{2} t + \sin^{2} t = 1$ verifies the ellipse.
At $t = 0$ , the point is $(2, 0)$ and its velocity is $(0, - 1)$ . The motion therefore moves downward from the rightmost point, giving a clockwise traversal.

Final answer

$x (t) = 2 \cos t$ , $y (t) = - \sin t$ , $0 \leq t \leq 2 π$ .

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Problem 9

Original problem · added worked solution

9. Original source equation or figure

The denominator vanishes at $x = - 1$ , inside the integration interval. An ordinary improper integral must converge separately on both sides of this pole.
Near $x = - 1$ , $x^{5} + 1 = 5 (x + 1) + O ((x + 1)^{2})$ , so the integrand has the same leading singularity as $\frac{1}{5 (x + 1)}$ .
The integral of that comparison expression tends to $- \infty$ from the left and $+ \infty$ from the right. Each one-sided integral fails to be finite, so the ordinary improper integral diverges. Opposite infinities cannot be cancelled in this definition.

Final answer

The improper integral diverges because of the pole at $x = - 1$ .

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Problem 10

Original problem · added worked solution

10.

Replace the infinite endpoint by a limit: $\int_{0}^{\infty} \frac{x}{x^{2} + 1} d x = \lim_{b \to \infty} \int_{0}^{b} \frac{x}{x^{2} + 1} d x$ .
Let $u = x^{2} + 1$ , so $d u = 2 x d x$ . The integral to $b$ is $\frac{1}{2} \ln (b^{2} + 1) - \frac{1}{2} \ln 1$ .
Because $\ln (b^{2} + 1) \to \infty$ , the finite-endpoint integrals increase without bound.

Final answer

Diverges to $+ \infty$ .

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Problem 11

Added problem and worked solution

Evaluate $\int \frac{1}{\sqrt{9 - x^{2}}} d x$ on $| x | < 3$ .

Let $x = 3 \sin θ$ with $- \frac{π}{2} < θ < \frac{π}{2}$ . Then $d x = 3 \cos θ d θ$ and $\sqrt{9 - x^{2}} = 3 \cos θ$ .
The factors cancel, leaving $\int d θ = θ + C$ .
Replace $θ$ with $\arcsin (\frac{x}{3})$ . The derivative of this expression is the original integrand for $| x | < 3$ .

Final answer

$\arcsin (\frac{x}{3}) + C$ .

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Problem 12

Added problem and worked solution

Evaluate $\int \frac{x^{2}}{\sqrt{4 - x^{2}}} d x$ on $| x | < 2$ .

Let $x = 2 \sin θ$ , so $d x = 2 \cos θ d θ$ and the radical is $2 \cos θ$ . The integral becomes $4 \int \sin^{2} θ d θ$ .
Use $\sin^{2} θ = \frac{1 - \cos (2 θ)}{2}$ to obtain $2 θ - \sin (2 θ) + C$ .
Since $\sin (2 θ) = \frac{x \sqrt{4 - x^{2}}}{2}$ , return to $x$ .

Final answer

$2 \arcsin (\frac{x}{2}) - \frac{x \sqrt{4 - x^{2}}}{2} + C$ .

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Problem 13

Added problem and worked solution

Evaluate $\int \sqrt{x^{2} + 4} d x$ .

Put $x = 2 \tan θ$ . Then the integral becomes $4 \int \sec^{3} θ d θ$ .
Integration by parts on $J = \int \sec^{3} θ d θ$ gives $J = \sec θ \tan θ - \int \sec θ \tan^{2} θ d θ$ . Replace $\tan^{2} θ$ by $\sec^{2} θ - 1$ and rearrange: $2 J = \sec θ \tan θ + \ln | \sec θ + \tan θ |$ .
Multiply by 4 and substitute $\tan θ = \frac{x}{2}$ and $\sec θ = \frac{\sqrt{x^{2} + 4}}{2}$ . A constant logarithmic factor is absorbed into $C$ .

Final answer

$\frac{x \sqrt{x^{2} + 4}}{2} + 2 \ln (x + \sqrt{x^{2} + 4}) + C$ .

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Problem 14

Added problem and worked solution

Evaluate $\int_{0}^{3} \frac{1}{(x^{2} + 9)^{3 / 2}} d x$ .

Let $x = 3 \tan θ$ , so $d x = 3 \sec^{2} θ d θ$ and the denominator is $27 \sec^{3} θ$ . The integrand becomes $\frac{1}{9} \cos θ d θ$ .
The endpoints $x = 0, 3$ correspond to $θ = 0, \frac{π}{4}$ .
Integrate: $\frac{1}{9} [\sin θ]_{0}^{π / 4} = \frac{\sqrt{2}}{18}$ .

Final answer

$\frac{\sqrt{2}}{18}$ .

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Problem 15

Added problem and worked solution

Evaluate $\int x \ln x d x$ for $x > 0$ .

Choose $u = \ln x$ and $d v = x d x$ , so $d u = \frac{1}{x} d x$ and $v = \frac{x^{2}}{2}$ .
Integration by parts gives $\frac{x^{2}}{2} \ln x - \frac{1}{2} \int x d x$ .
Integrate the remaining power to obtain the result on $x > 0$ .

Final answer

$\frac{x^{2}}{2} \ln x - \frac{x^{2}}{4} + C$ .

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Problem 16

Added problem and worked solution

Evaluate $\int x^{2} \cos x d x$ .

Use $u = x^{2}$ and $d v = \cos x d x$ to obtain $x^{2} \sin x - 2 \int x \sin x d x$ .
For the remaining integral use $u = x$ , $d v = \sin x d x$ : $\int x \sin x d x = - x \cos x + \sin x$ .
Substitute this expression and collect terms.

Final answer

$x^{2} \sin x + 2 x \cos x - 2 \sin x + C$ .

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Problem 17

Added problem and worked solution

Evaluate $\int e^{x} \sin x d x$ .

Let $I = \int e^{x} \sin x d x$ . With $u = \sin x$ and $d v = e^{x} d x$ , $I = e^{x} \sin x - J$ , where $J = \int e^{x} \cos x d x$ .
Integrating $J$ by parts gives $J = e^{x} \cos x + I$ .
Substitute into the first equation: $2 I = e^{x} (\sin x - \cos x)$ . Divide by 2 and include $C$ .

Final answer

$\frac{e^{x}}{2} (\sin x - \cos x) + C$ .

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Problem 18

Added problem and worked solution

Evaluate $\int_{1}^{e} \ln x d x$ .

Let $u = \ln x$ and $d v = d x$ . Then $d u = \frac{1}{x} d x$ and $v = x$ .
An antiderivative is $x \ln x - x$ .
Evaluate at $e$ and $1$ : $(e - e) - (0 - 1) = 1$ .

Final answer

$1$ .

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Problem 19

Added problem and worked solution

Evaluate $\int \frac{3 x + 5}{x^{2} + x - 2} d x$ on intervals avoiding the poles.

Factor $x^{2} + x - 2 = (x - 1) (x + 2)$ . Write $\frac{3 x + 5}{(x - 1) (x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}$ .
Equate coefficients in $3 x + 5 = A (x + 2) + B (x - 1)$ : $A + B = 3$ and $2 A - B = 5$ . Thus $A = \frac{8}{3}$ and $B = \frac{1}{3}$ .
Integrate both reciprocal linear terms with logarithms.

Final answer

$\frac{8}{3} \ln | x - 1 | + \frac{1}{3} \ln | x + 2 | + C$ .

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Problem 20

Added problem and worked solution

Evaluate $\int \frac{1}{x (x + 1)^{2}} d x$ for $x \neq 0, - 1$ .

Write $\frac{1}{x (x + 1)^{2}} = \frac{A}{x} + \frac{B}{x + 1} + \frac{D}{(x + 1)^{2}}$ .
Multiplying by the denominator gives $1 = A (x + 1)^{2} + B x (x + 1) + D x$ . Substituting $x = 0$ and $x = - 1$ yields $A = 1$ , $D = - 1$ ; the quadratic coefficient gives $B = - 1$ .
Integrate $\frac{1}{x} - \frac{1}{x + 1} - \frac{1}{(x + 1)^{2}}$ term by term.

Final answer

$\ln | x | - \ln | x + 1 | + \frac{1}{x + 1} + C$ .

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Problem 21

Added problem and worked solution

Evaluate $\int \frac{x^{3}}{x^{2} + 1} d x$ .

Divide the numerator by the denominator: $\frac{x^{3}}{x^{2} + 1} = x - \frac{x}{x^{2} + 1}$ .
The first integral is $\frac{x^{2}}{2}$ . For the second use $u = x^{2} + 1$ , $d u = 2 x d x$ .
Subtract the resulting logarithmic term.

Final answer

$\frac{x^{2}}{2} - \frac{1}{2} \ln (x^{2} + 1) + C$ .

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Problem 22

Added problem and worked solution

Evaluate $\int \frac{x + 1}{(x - 1) (x^{2} + 1)} d x$ for $x \neq 1$ .

Set $\frac{x + 1}{(x - 1) (x^{2} + 1)} = \frac{A}{x - 1} + \frac{B x + D}{x^{2} + 1}$ .
Multiplication gives $x + 1 = A (x^{2} + 1) + (B x + D) (x - 1)$ . Comparing coefficients gives $A = 1$ , $B = - 1$ , $D = 0$ .
Integrate $\frac{1}{x - 1} - \frac{x}{x^{2} + 1}$ , using logarithms for both terms.

Final answer

$\ln | x - 1 | - \frac{1}{2} \ln (x^{2} + 1) + C$ .

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Problem 23

Added problem and worked solution

Eliminate the parameter from $x = 2 t + 1$ , $y = t^{2}$ , $- \infty < t < \infty$ , and describe the direction of travel.

From $x = 2 t + 1$ , solve $t = \frac{x - 1}{2}$ .
Substitute into $y = t^{2}$ to obtain $y = \frac{(x - 1)^{2}}{4}$ .
As $t$ increases, $x$ increases because $\frac{d x}{d t} = 2 > 0$ ; the motion travels from left to right along the parabola.

Final answer

$y = \frac{(x - 1)^{2}}{4}$ , traced from left to right.

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Problem 24

Added problem and worked solution

Find the tangent line to $x = t^{2}$ , $y = t^{3} + t$ at $t = 1$ .

At $t = 1$ , the point is $(1, 2)$ .
Compute $\frac{d x}{d t} = 2 t$ and $\frac{d y}{d t} = 3 t^{2} + 1$ . Since $\frac{d x}{d t} \neq 0$ at $t = 1$ , $\frac{d y}{d x} = \frac{3 t^{2} + 1}{2 t} = 2$ there.
Use point-slope form through $(1, 2)$ .

Final answer

$y - 2 = 2 (x - 1)$ , or $y = 2 x$ .

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Problem 25

Added problem and worked solution

Find the arclength of $x = t^{2}$ , $y = \frac{2}{3} t^{3}$ for $0 \leq t \leq 1$ .

Differentiate: $x' = 2 t$ , $y' = 2 t^{2}$ . On $0 \leq t \leq 1$ , the speed is $2 t \sqrt{1 + t^{2}}$ .
Set $u = 1 + t^{2}$ , $d u = 2 t d t$ . The arclength is $\int_{1}^{2} \sqrt{u} d u$ .
Evaluate $\frac{2}{3} [u^{3 / 2}]_{1}^{2}$ .

Final answer

$\frac{2}{3} (2 \sqrt{2} - 1)$ .

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Problem 26

Added problem and worked solution

Parametrize $\frac{(x - 1)^{2}}{9} + \frac{(y + 2)^{2}}{4} = 1$ counterclockwise, beginning at its rightmost point.

The horizontal and vertical semiaxes are $3$ and $2$ , with center $(1, - 2)$ .
Choose $x = 1 + 3 \cos t$ , $y = - 2 + 2 \sin t$ . At $t = 0$ the point is $(4, - 2)$ and the velocity points upward.
Thus the traversal is counterclockwise, and substitution verifies the displayed ellipse.

Final answer

$x = 1 + 3 \cos t$ , $y = - 2 + 2 \sin t$ , $0 \leq t \leq 2 π$ .

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Problem 27

Added problem and worked solution

Determine convergence and evaluate $\int_{1}^{\infty} \frac{1}{x^{3}} d x$ if it converges.

Write the integral as $\lim_{b \to \infty} \int_{1}^{b} x^{- 3} d x$ .
An antiderivative is $- \frac{1}{2 x^{2}}$ , so the finite integral equals $\frac{1}{2} - \frac{1}{2 b^{2}}$ .
The second term tends to zero, so the limit is finite.

Final answer

Converges to $\frac{1}{2}$ .

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Problem 28

Added problem and worked solution

Determine convergence and evaluate $\int_{0}^{1} \frac{1}{\sqrt{x}} d x$ if it converges.

Replace the singular lower endpoint by $ε > 0$ : $\lim_{ε \to 0^{+}} \int_{ε}^{1} x^{- 1 / 2} d x$ .
Integrate to get $2 - 2 \sqrt{ε}$ .
Since $\sqrt{ε} \to 0$ , the improper integral has a finite limit.

Final answer

Converges to $2$ .

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Problem 29

Added problem and worked solution

Determine convergence and evaluate $\int_{0}^{1} \ln x d x$ if it converges.

Integrate from $ε$ to 1. Integration by parts gives the antiderivative $x \ln x - x$ .
The finite integral is $- 1 - ε \ln ε + ε$ . To evaluate the endpoint limit, put $ε = e^{- t}$ ; then $ε \ln ε = - t e^{- t} \to 0$ .
Taking $ε \to 0^{+}$ gives a finite result.

Final answer

Converges to $- 1$ .

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Problem 30

Added problem and worked solution

Does the ordinary improper integral $\int_{0}^{2} \frac{1}{x - 1} d x$ converge? Explain using its one-sided limits.

The integrand has a pole at $x = 1$ inside the interval, so split the improper integral into separate limits on $(0, 1)$ and $(1, 2)$ .
On the left, $\int_{0}^{1 - ε} \frac{1}{x - 1} d x = \ln ε \to - \infty$ . On the right, $\int_{1 + ε}^{2} \frac{1}{x - 1} d x = - \ln ε \to + \infty$ .
Both one-sided integrals must be finite for the ordinary improper integral to converge. They are not, even though symmetric truncations cancel.

Final answer

The ordinary improper integral diverges.

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