Category Archives: Introductory Problems

Numerical Sequences and Series: The comparison test

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Determine if the series, a1 + a2 + a3 + a4 + … converges or diverges, where:

1. an= 1/(3n + 2)

Answer

1. Converges

Solution

1. First, observe that 0 < 1/(3n + 2) < 1/3n The series with bn = 1/3n is a geometric series with r= 1/3 and so converges. Then the series with an= 1/(3n + 2) converges by the comparison test.

 

2. an= 1/(n – 1)

Answer

1. Diverges

Solution

1. First, observe that 1/(n – 1) > 1/n The series with bn= 1/n is the harmonic series and so diverges. Then the series with an = 1/(n – 1) diverges by the comparison test.

 

3. an = 1/(n2 + n)

Answer

1. Converges

Solution

1. This is a little trickier, because we could compare this an with 1/n (the harmonic series, so diverges) or with 1/n2 (a p-series with p = 2 > 1, so converges). Which should be used?
Note: 1/(n2 + n) < 1/n and 1/(n2 + n) < 1/n2 Being less than a diverging series tells us nothing, but being less than a converging series tells us that the series of an converges.

 

4. an = 1/(3n – 2)

Answer

1. Converges

Solution

1. The Comparison Test cannot be applied, because 1/(3n – 2) > 1/3n and although the geometric series ∑ 1/3n converges, being greater than a converging series tells us nothing. We can use the Limit Comparison Test, because as n → ∞, (1/(3n – 2)/(1/3n) = (3n)/(3n – 2) = 1/(1 – 2/3n) → 1 Then by the Limit Comparison Test, ∑ 1/(3n – 2) converges because ∑ 1/3n converges.

 

5. an = (√(n3 + 4))/(n2 + 2n + 3)

Answer

1. Diverges

Solution

1. The first issue is to find what series to use for comparison. For large n the numerator √(n3 + 4) is close to n3/2. For large n the denominator n2 + 2n + 3 is close to n2. Then the fraction an is close to n3/2/n2 = 1/n1/2. This is the general term of the series we use for comparison. As n → ∞, (√(n3 + 4))/(n2 + 2n + 3) / (1/n1/2) = (n3/2 √(1 + 4/n3)) / (n2(1 + 2/n + 3/n2)) ⋅ n1/2 = (√(1 + 4/n3)) / (1 + 2/n + 3/n2) → 1 Being a p-series with p = 1/2 < 1, ∑1/n1/2 diverges, so the original series diverges by the Limit Comparison Test.

 

6. an = (3n + 2)/(5n + 4)

Answer

1. Converges

Solution

1. For large n the ratio (3n + 2)/(5n + 4) is very close to 3n/5n, the general term of a geometric series with ratio 3/5, hence converging. As n → ∞, ((3n + 2)/(5n + 4))/(3n/5n) = ((3n(1 + 2/3n))/(5n(1 + 4/5n))⋅(5n/3n) = (1 + 2/3n)/(1 + 4/5n) → 1 Then by the Limit Comparison Test, the original series converges because the geometric series converges.

 

 

Numerical Sequences and Series: The integral test

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Determine if the series, a1 + a2 + a3 + a4 +… converges or diverges, when:

1. an = n/(n2 + 1)

Answer

1. Diverges

Solution

1. f(x)= x/(x2 + 1) is positive, continuous, and decreasing because f'(x)= (1 – x2)/(1 + x2)2 is negative for x > 1. Substituting u= x2 + 1 so du= 2x dx and x dx = (1/2)du, we see ∫x/(x2 + 1) dx = (1/2) ∫du/u = (1/2)ln(u) = (1/2)ln(x2 + 1) This diverges to ∞ as x → ∞, so the series diverges by the Integral Test.

 

2.an= n2/(n3 + 1)

Answer

1. Diverges

Solution

1. f(x)= x2/(x3 + 1) is positive, continuous. Now f'(x)= x(2 – x3)/(1 + x3)2 is negative for x > 21/3≈ 1.2599, so we can apply the Integral test only to the series starting with n= 2. This is fine: the n = 1 term cannot affect the convergence of the series. Substituting u= x3 + 1, so du = 3x2 dx and x2 dx = (1/3)du, we see ∫x2/(x3 + 1) dx = (1/3) ∫du/u = (1/3) ln(u) = (1/3) ln(x3 + 1) This diverges to ∞ as x → ∞, so the series diverges by the Integral Test.

 

3. an= n/(n2 + 1)2

Answer

1. Converges

Solution

1. f(x)= x/(x2 + 1)2 is positive, continuous, and decreasing because f'(x)= (1 – 3x2)/(1 + x2)3is negative for x > 1. Substituting u = x2 + 1 so du = 2x dx and x dx = (1/2)du, we see: ∫x/(x2 + 1)2 dx = (1/2)∫ du/u2 = -(1/2)(1/u) = -(1/2)(1/(x2 + 1)) This converges, so the series converges by the Integral Test.

 

4. an=1/(n ln(n)), starting with n = 2

Answer

1. Diverges

Solution

1. f(x)= 1/(x ln(x)) is positive, continuous, and decreasing because f'(x)= -(1 + ln(x))/(x2ln2(x)) is negative for > 1. Substituting u= ln(x) so du = (1/x)dx, we see ∫1/(x ln(x)) dx = ∫ du/u = ln(u) = ln(ln(x)) This diverges to ∞ as x → ∞, so the series diverges by the Integral Test.

 

Numerical Sequences and Series: Zeno’s Paradox

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1. Find the sum of the series 1 – 1/2 + 1/4 – 1/8 + …

Answer

1. The series sums to 2/3.

Solution

 

2. Find the sum of the series 3 + 3/2 + 3/4 + 3/8 + …

Answer

1. The series sums to 6.

Solution

1. Factoring out a 3, this series is 3⋅(1 + 1/2 + 1/4 + 1/8 + …) The series in parentheses is a geometric series with ratio r = 1/2, converging because |r| < 1. This series sums to 1/(1 – r) = 1/(1 – (1/2)) = 2, so the original series sums to 3⋅2 = 6

 

3. Find the sum of the series 1/9 + 1/27 + 1/81 + 1/243 + …

Answer

1. The series sums to 1/6.

Solution

1. Factoring out a 1/9 from every term, we see the series is (1/9)⋅(1 + 1/3 + 1/9 + 1/27 + …) The series in parentheses is a geometric series with ratio r = 1/3, converging because |r| < 1. This series sums to 1/(1 – r) = 3/2, so the original series sums to (1/9)⋅(3/2) = 1/6

 

4. Find the sum of the series 1 + 2 + 4 + 8 + 16 + … + 128

Answer

1. The series sums to 255

Solution

1. Recall for all r, 1 + r + r2 + … + rN-1= (1 – rN)/(1 – r) Because 128= 27, we see that 1 + 2 + 4 + … + 128 = (1 – 28)/(1 – 2) = (1 – 256)/(1-2) = -255.

 

5. Generalize problem 4 to 1 + 2 + 4 + … + 2n

Answer

1. The series sums to 2n+1– 1

Solution

1. Recall for all r, 1 + r + r2 + … + rN-1= (1 – rN)/(1 – r) Then 1 + 2 + 4 + … + 2n= (1 – 2n+1)/(1 – 2)= 2n+1– 1.

 

Numerical Sequences and Series

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In problems 1 – 4 determine if the sequences converge or diverge. For those that converge, find the limits.

 

1. an= (n2 + 3)/(2n2 – 5)

Answer

1. Converges to 1/2.

Solutions

2. Factoring n2 from numerator and denominator, we find limn→ ∞ (n2 + 3)/(2n2 – 5) = limn<→ ∞(1 + 3/n2)/(2 – 5/n2) = (1 + 0)/(2 – 0) = 1/2

 

2. an = (n2 + 3)/(2n3 – 5)

Answer

1. Converges to 0.

Solution

1. Factoring n3 from numerator and denominator, we find limn → ∞(n2 + 3)/(2n3 – 5)= limn → ∞(1/n + 3/n3)/(2 – 5/n3) = (0 + 0)/(2 – 0) = 0/2 = 0

 

3. an = (n2 + 3)/(2n – 5)

Answer

1. Diverges

Solution

1. Factoring n from numerator and denominator, we find limn → ∞ (n2 + 3)/(2n – 5)= limn → ∞(n + 3/n)/(2 – 5/n) =limn → ∞ n/2= ∞

 

4. an = √(n+1) – √n

Answer

1. Converges to 0

Solution

1. Multiply by (√(n+1) + √n)/(√(n+1) + √n), obtaining limn → ∞ √(n+1) – √n = limn → ∞ (√(n+1) – √n)(√(n+1) + √n)/(√(n+1) + √n) = limn → ∞ (n + 1 – n)/(√(n+1) + √n) = limn → ∞ 1/(√(n+1) + √n) = 0

 

In problems 5 – 8 determine if the series ∑n=1∞(a)n converge or diverge. For those that converge, find the limits. In problem 6, start the sum at n = 2.

5. an = n2/(2n2 + 1)

Answer

1. Diverges

Solution

1. Because limn → ∞ n2/(2n2 + 1)= limn → ∞ 1/(2 + 1/n2) = 1/(2 + 0) = 1/2 the series diverges by the 9th term test.

 

6. an = 2/(n2 – 1)

Answer

1. Converges to 3/2

Solution

1. Apply the method of partial fractions and conclude: 2/(n2 – 1) = 1/(n – 1) – 1/(n + 1) Starting the series from n = 2, the first few terms are (1/(2-1) – 1/(2+1)) + (1/(3-1) – 1/(3+1)) + (1/(4-1) – 1/(4+1)) + (1/(5-1) – 1/(5+1)) + … = (1 – 1/3) + (1/2 – 1/4) + (1/3 – 1/5) + (1/4 – 1/6) + … = 1 + 1/2 = 3/2 because 1/3 occurs twice, once + and once -, 1/4 occurs twice, once + and once -, and so on. Only the 1 and 1/2 are not cancelled by later terms.

 

7. an = n sin(1/n)

Answer

1. Diverges

Solution

1. Note that n sin(1/n) = sin(1/n)/(1/n). Because 1/n → 0 as n → ∞, and recalling limx → 0 sin(x)/x = 1, we see limn → ∞n sin(1/n) = 1. The series diverges by the nth term test.

 

8. an = arctan(n)

Answer

1. Diverges

Solution

1. As n → ∞, arctan(n) → π/2, so the series diverges by the nth term test.

 

 

 

 

Complex numbers and Euler’s formula

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1. Evaluate (a)(2 + 3i) + (4 – 7i) (b) (2 + 3i) + (4 – 7i)

Answer

    1. 1. (a) 6 – 4i (b) -2 + 10i

 

2. Evaluate (2 + 3i)⋅(4 – 7i) (2 + 3i)2

Answer

    1. (a) 29 – 2i (b) -5 + 12i

 

3. Evaluate (2 + 3i)/(4 – 7i)

Answer

    1. -1/5 + (2/5)i

Solution

    1. (2 + 3i)/(4 – 7i) = (2 + 3i)/(4 – 7i) ⋅ (4 + 7i)/(4 + 7i) = (2 + 3i)⋅(4 + 7i) / (42 + 72) =                                (-13 + 26i)/65 = -1/5 + (2/5)i

 

4. Find the square roots of 9 – 9i.

Answer

    1. 3⋅21/4 (cos(3π/8) + i sin(3π/8) and -3 ⋅21/4 (cos(3π/8) + i sin(3π/8)

Solution

    1. 9 – 9i has modulus (92 + 92)1/2 = 9√2, and argument 3π/4. Then one square root has             modulus (9√2)1/2 = 3⋅21/4 and argument 3π/8, so is 3⋅21/4(cos(3π/8) + i sin(3π/8)) ≈                 1.365 + 3.296i. The other square root is -3⋅21/4(cos(3π/8) + i sin(3π/8)) ≈ -1.365 –                     3.296i.

 

5. Solve z2 = 9 – 9i.

Answer

    1. 3⋅21/4 (cos(3π/8) + i sin (3π/8)) and -3⋅21/4 (cos(3π/8) + i sin (3π/8)

Solution

    1. The solutions are z = square roots of 9 + 9i, solved in problem 4

 

 

Area of surfaces of revolution

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1. Find the area of the surface obtained by revolving y = sin(x), 0 ≤ x ≤ π, about the x-axis.

Answer

    1. The area is 2π(√2 + (1/2)ln(1 + √2) – (1/2)ln(-1 + √2))

Solution

    1. Because f(x) = sin(x), f'(x) = cos(x) and the area is ∫0π2π sin(x) √(1 + cos2(x)) dx = -2π∫1-           1√(1 + u2)du substituting u = cos(x) = -2π((u/2)√(1 + u2) + (1/2)ln(u + √(1 + u2)))|1-1 by           16 from the Integral Table = 2π(√2 + (1/2)ln(1 + √2) – (1/2)ln(-1 + √2))

 

2. Find the area of the surface obtained by revolving y = √x, 0 ≤ x ≤ 1, about the x-axis.

Answer

    1. The area is (5√5 – 1)/12.

Solution

    1. Because f(x) = √x, f ‘(x) = 1/(2√x) and the area is ∫01 2π √x √(1 + (1/(2√x))2) dx = 2π∫01√(x           + 1/4) dx = 2π (2/3)(x + 1/4)3/2|01 = (5√5 – 1)/12

 

3. Find the area of the surface obtained by revolving y = ex, 0 ≤ x ≤ 1, about the x-axis.

Answer

    1. The area is -2π (((1/2e)√(1 + 1/e2) + (1/2)ln(1/e + √(1 + 1/e2)) – (√2/2 + (1/2)ln(1 + √2)))

Answer

    1.  Because f(x) = e-x, f'(x) = -e-x and the area is ∫01 2π e-x √(1 + (-e-x)2) dx = -2π∫11/e√(1 +              u2) du substituting u = e-x = -2π((u/2)√(1 + u2) + (1/2)ln(u + √(1 + u2)))|11/e by 16 from            the Integral Table = -2π (((1/2e)√(1 + 1/e2) + (1/2)ln(1/e + √(1 + 1/e2)) – (√2/2 + (1/2) ln(1          + √2)))

 

4. Find the area of the surface obtained by revolving y = √(x+1), 0 ≤ x ≤ 1, about the x-axis.

Answer

    1. The area is π(27 – 5√5)/6.

Solution

1. Because f(x) = √(x+1), f ‘(x) = 1/(2√(x+1)) and the area is ∫01 2π √(x+1) √(1 + (1/(2√(x+1)2) dx = 2π∫01√(x + 1 + 1/4) dx = 2π (2/3)(x + 5/4)3/2|01 = π(27 – 5√5)/6

 

5. Find the area of the surface obtained by revolving y = √(2x – x2), 0 ≤ x ≤ 1, about the x-axis.

Answer

1. The area is 2π.

Solution

1. Because f(x) = √(2x – x2), f ‘(x) = (1 – x)/√(2x – x2) and the area is ∫01 2π √(2x – x2) √(1 + ((1 – x)/√(2x – x2)2) dx = 2π∫01 √(2x – x2 + (1 – x)2) dx = 2π∫01 √1 dx = 2π

 

 

Arclength

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1. Find the length of f(x) = mx from x = 0 to x = 1. Does this agree with basic geometry?

Answer

    1. Length = √(1 + m2)

Solution

    1. First, f'(x) = m, so √(1 + (f'(x))2) = √(1 + m2). The arclength is ∫01√(1 + m2) dx = √(1 + m2)x|01= √(1 + m2). This agrees with basic geometry, because the path is the straight line from (0, 0) to (1, m), the hypotenuse of a right triangle with base 1 and height m.

 

2. Find the length of r(t) = < cos(t), sin(t) > from t = 0 to t = π/2.

Answer

    1. The length is π/2.

Solution

    1. First, note r'(t) = < -sin(t), cos(t) >, so |r'(t)| = √(sin2(t) + cos2(t)) = 1. The arclength is:               ∫0π/2 1 dt = t|0π/2 = π/2

 

3. Find the length of r(t) = < cos(t2), sin(t2) > from t = 0 to t = π/2.

Answer

    1. Length = π2/4.

Solution

    1. First, note r'(t) = < -sin(t2)2t, cos(t2)2t >, so |r'(t)| = √(sin2(t2)4t2 + cos2(t2)4t2) = 2t. The arclength is ∫0π/2 2t dt = t2|0π/2= π2/4

 

4. Find the length of f(x) = x2/2 from x = 0 to x = 1.

Answer

1. The arclength is (√2)/2 + ln(1 + √2)/2.

Solution

    1. First, f'(x) = x, so √(1 + (f'(x))2) = √(1 + x2). Applying formula 16 from the integral table, we find the arclength is (√2)/2 + ln(1 + √2)/2.

 

5. Find the length of f(x) = (2/3)x3/2 from x = 0 to x = 1

Answer

    1. The arclength is (4√2 – 2)/3.

Solution

    1. First, f'(x) = x1/2, so √(1 + (f'(x))2) = √(1 + x) and the arclength is: ∫0π/2√(1 + x) dx = (2/3)(1 +           x)3/2|01= (4√2 – 2)/3.

 

 

 

Parametric curves

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1. For the curve x(t) = t2 and y(t) = 3t – 1, write y as an explicit function of x

Answer

1. y= 3x1/2 – 1

Solution

1. Solve x(t) = t2 for t, obtaining t = x1/2. Then substitute this in for t in y(t) = 3t – 1: y = 3x1/2 -1

 

2. Plot x(t) = sin(at), y(t) = sin(bt) for (a,b) = (2,1), (2,2), and (2,3).

Answer

1. (a,b) = (2,1) 2. (a,b) = (2,2) 3. (a,b) = (2,3)

Solution

1.  The solutions are the following: For (a,b) = (2,1), the x-coordinate makes two complete circuits, the y-coordinate one. The curve passes through the origin so is a figure 8. For (a,b) = (2,2), the x- and y-coordinates are the same, so the curve is the line of slope 1, -1 ≤ x ≤ 1, through the origin. For (a,b)= (2,3) the x-coordinate makes two complete circuits, the y-coordinate three. Again, the curve passes through the origin.

 

3. For the curve x(t) = t2 + 2t – 1, y(t) = t3 + t2 – 5t, find the points (x,y) at which the tangent is horizontal; find the points at which the tangent is vertical.

Answer

1. The horizontal tangents occur at (x, y) = (175/27, 4/9) and (x, y) = (-3, 4). The vertical tangents occur at (x, y) = (5, 0).

Solution 

1. The slope of the tangent line is dy/dx = (t3 + t2 – 5t)’/(t2 + 2t + 1)’ = (3t2 + 2t – 5)/(2t + 2). The tangent line is horizontal when 3t2 + 2t – 5 = 0. That is, t = -5/3 and t = 1. The corresponding points are (x, y) = (175/27, 4/9) and (x, y) = (-3, 4). The tangent line is vertical when 2t + 2 = 0. That is, t = -1. The corresponding point is (x, y) = (5, 0).

 

4. Describe the curve x(t) = 3 cos(t), y(t) = 3 sin(t), z(t) = 3 cos(t), for 0 ≤ t ≤ 2π.

Answer

1. The curve is an ellipse lying in the plane z = x, with semi-minor axis length 3 and semi-major axis length 3√2.

Solution

1. The curve is the graph of the cosine function in the plane y = x.

 

5. Describe the curve x(t) = t, y(t) = t, z(t) = cos(t), for 0 ≤ t ≤ 2π.

Answer

1. The curve is the graph of the cosine function in the plane y = x.

Solution

1. Because x = t and y = t, the curve lies in the plane y = x. Projected into the xz plane, the curve is z = cos(t) = cos(x). Similarly, in the yz plane the curve projects to z = cos(y). Then the curve is the graph of the cosine function, but stretched horizontally, because successive maxima of the curve are (0,0,1) and (2π,2π,1). That is, the distance between successive maxima is 2π√2.

 

 

Volumes by cross-sections

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1. Find the volume of the solid with right isosceles triangular cross-section perpendicular to the x-axis, with base x2, for 0 ≤ x ≤ 1

Answer

    1. Volume = 1/10.

Solution

     1. A right isosceles triangle with base x2 has altitude x2 and so area A(x) = (1/2)⋅base⋅altitude (1/2)⋅x2⋅x2 = (1/2)⋅x4 Then the volume is: V = ∫01A(x) dx = ∫01(1/2)⋅x4dx = x5/10|01= 1/10.

 

2. Find the volume if the solid with elliptical cross-section perpendicular to the x-axis, with semi-major axis x2 and semi-minor axis x3, for 0 ≤ x ≤ 1

Answer

    1. Volume = π/6.

Solution

    1. Recall an ellipse with semi-major axis a and semi-minor axis b has area πab, so this ellipse with semi-major axis x2 and semi-minor axis x3 has the area: A(x) = π⋅x2⋅x3 = π⋅x5. Then the volume is V= ∫01A(x) dx = ∫01π⋅x5dx = π⋅x6/6|01 = π/6.

 

3. Find the volume of the solid with circular cross-section of radius cos3/2(x), for 0 ≤ x ≤ π/2.

Answer

    1. Volume = 2π/3.

Solutions

    1. The area is A(x) = π(cos3/2(x)2 =πcos3(x). The volume is V = ∫0π/2A(x) dx = ∫0π/2πcos3(x) dx= π∫0π/2cos(x)(1 – sin2(x)) dx = π∫0π/2cos(x)dx – π∫0π/2cos(x)sin2(x) dx = πsin(x)|0π/2 – πsin3(x)/3= |0π/2 = π – π/3 = 2π/3. where cos(x)sin2(x) is integrated using the substitution u = sin(x), so du = cos(x) dx.

 

4. Find the volume of the solid with cross-section a rectangle of base x and height ex2

Answer

    1. Volume= (e – 1)/2.

Solution

    1. The area is A(x) = base ⋅ height = x⋅ex2. The volume is V = ∫01A(x) dx = ∫01 x⋅ex2 dx= (ex2)/2|01 = (e – 1)/2. where x⋅ex2 was integrated using the substitution u = x2, so du = 2xdx.

 

5. Find the volume of the solid obtained by rotating the curve y = x2, -1 ≤ x ≤ 1, about the x-axis.

Answer

    1. Volume = 2π/5.

Solution

    1. The cross-sections are circles of radius x2, so the cross-sectional area is A(x) π⋅(x2)2π⋅x4 The volume is V = ∫-11A(x) dx = ∫-11 π⋅x4 dx = π⋅(x5/5)|-11 = 2π/5

 

 

Using integration tables

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More Challenging Problems 

 

1. Find ∫ [2 + 3x2]-1/2dx

Answer

    1. 3-1/2 ln |x + √ (2/3 + x2)| + C

Solution

    1. The closest formula is number 20. from the notes, except for the “3” in front of x2. We can            pull it out of the parentheses: [2 + 3x2]-1/2= 3-1/2 [2/3 + x2]-1/2 Now we can use the formula,            with a = √ (2/3): ∫[2 + 3x2]-1/2 dx = ∫ 3-1/2 [2/3 + x2>]-1/2 dx = 3-1/2 ln | x + √ (2/3 + x2) | + C

 

2. Find ∫ sin3(x) dx

Answer

    1. -1/3 sin2(x) cos(x) – 2/3 cos(x) + c

Solution

     1. We can use formula 7. from the notes, with n = 2: ∫sin3(x) dx = – 1/3 sin2(x) cos(x) + 2/3 ∫                   sin(x) dx = – 1/3 sin2(x) cos(x) – 2/3 cos(x) + c

 

3. Find ∫ tan2(2x) dx

Answer

1. 1/2 tan (2x) – x + c

Solution

    1. The notes have formula 3. for tan2(x). To use it, first substitute u = 2x, so du = 2 dx, and we            get: ∫tan2(2x) dx = ∫ 1/2 tan2(u) du = 1/2 [ tan(u) – u ] + c= 1/2 tan (2x) -x + c

 

4. Find ∫2x arccos(x2) dx

Answer

    1. x2 arccos(x2)- √(1 – x4) + C

Solution

    1. Formula 47. in the notes has arccos (or cos-1) in it. We can’t use it as stated, since we need x2 inside the arccos, and we have an extra 2x up front. This is an ideal case for substitution, since 2x is the derivative of x2. Use u = x2, so du = 2x dx: ∫2x arccos(x2) dx = ∫ arccos(u) du. Now we use formula 47.: int; arccos(u) du = u arccos(u) – √ ( 1 – u2 ) + C Substituting back x2 for u, we get the answer: ∫ 2x arccos(x2) dx = x2 arccos(x2) – √(1 – x4) + C

 

5. Find ∫[x2 + 2x + 2]-1 dx

Answer

    1. Arctan(x + 1) + c

Solution

1. This does not look like anything in the tables. We have formulas for things such as a2 + x2, but      not summand of “x” in there. We have to put our expression into that form first, by absorbing the “2x” into the square: x2 + 2x + 2 = (x2 + 2x + 1) + 1 = (x + 1)2 + 1 With this, we have: ∫[x2 + 2x + 2]-1 dx = ∫[( x + 1)2 + 1]-1 dx This is close to formula 41. in the notes, if we substitute u = x + 1, du = dx: ∫[x2 + 2x + 2]-1 dx = ∫[(x + 1)2 + 1]-1 dx = ∫[u2 + 1]-1 du = arctan(u) + c = arctan(x + 1) + c