Integration Exercises with Solutions.pdf PDF

Preview

Summary

MATH 105 921 Solutions to Integration Exercises MATH 105 921 Solutions to Integration Exercises s2 + 1 Z 1) ds s2 − 1 Solution: Performing polynomial long division, we have that: Z 2 Z s +1 2 ds = (1 + ) ds s2 − 1 s2 − 1 Z Z 2 = ds + 2 ds s −1 Z 2 =s+ 2 ds s −1 Using partial fraction on the remainin...

Description

MATH 105 921

Solutions to Integration Exercises

MATH 105 921 Solutions to Integration Exercises 1)

s2 + 1 ds s2 − 1

Z

Solution: Performing polynomial long division, we have that: Z Z 2 2 s +1 ds = (1 + ) ds s2 − 1 s2 − 1 Z Z 2 ds = ds + 2 s −1 Z 2 =s+ ds 2 s −1 Using partial fraction on the remaining integral, we get: s2

2 A B A(s + 1) + B(s − 1) (A + B)s + (A − B) = + = = −1 s−1 s+1 (s + 1)(s − 1) s2 − 1

Thus, A + B = 0 and A − B = 2. Adding the two equations together yields 2A = 2, that is, A = 1, and B = −1. So, we have that: Z Z Z 1 1 2 ds = ds − ds 2 s −1 s−1 s+1 Therefore, Z

2)

Z

0

s2 + 1 ds = s + s2 − 1

2 ds s2 − 1 Z Z 1 1 =s+ ds − ds s−1 s+1 = s + ln |s − 1| − ln |s + 1| + C Z

√ x 1 + 2x dx

4

Solution: Using direct substitution with u = 1 + 2x and du = 2dx, we may write

Page 1 of 22

MATH 105 921

Solutions to Integration Exercises

x = 21 (u − 1). Moreover, when x = 4, u = 9, and when x = 0, u = 1. Thus, Z

3)

Z

0

√

1

√ 1 (u − 1) u du 4 Z9 1 1 1 3 (u 2 − u 2 ) du = 9 4 1 5 1 3 = ( u 2 − u 2 ) |19 10 6 1 1 243 27 =( − )−( − ) 10 6 10 6 −298 = 15

x 1 + 2x dx = 4

Z

sin2 x cos2 x dx

Solution: Using half-angle identities Z Z 2 2 sin x cos x dx = Z = Z =

sin2 x =

1−cos(2x) 2

and cos2 x =

1+cos(2x) , 2

we get:

1 (1 − cos(2x))(1 + cos(2x)) dx 4 1 (1 − cos2 (2x)) dx 4 Z 1 1 dx − cos2 (2x) dx 4 4 Z x 1 cos2 (2x) dx = − 4 4

On the remaining integral, we apply the half-angle identity cos2 (2x) = obtain: Z Z x 1 1 + cos(4x) 2 dx = + sin(4x) + C cos (2x) dx = 2 2 8

1+cos(4x) , 2

Hence, Z

4)

Z

sin2 x cos2 x dx =

x 1 x 1 x 1 − ( + sin(4x)) + C = − sin(4x) + C 4 4 2 8 8 32

√ sin( w) dw

Page 2 of 22

and

MATH 105 921

Solutions to Integration Exercises

√ Solution: Using direct substitution with t = w, and dt = √ 2 w dt = 2t dt, we get: Z Z √ sin( w) dw = 2t sin t dt

1 √ dw, 2 w

that is, dw =

Using integration by part method with u = 2t and dv = sin t dt, so du = 2 dt and v = − cos t, we get: Z Z 2t sin t dt = −2t cos t + 2 cos t dt = −2t cos t + 2 sin t + C Therefore, Z

5)

Z

√ √ √ √ sin( w) dw = −2 w cos( w) + 2 sin( w) + C

ln(x) dx x Solution: Using direct substitution with u = ln(x) and du =

1 x

dx, we get:

Z ln(x) u2 dx = u du = +C x 2 Z ln(x) 1 ⇒ dx = (ln(x))2 + C x 2 Z

6)

Z

sin t cos(2t) dt Solution: Recall the double-angle formula that cos(2t) = 2 cos2 t − 1, we get: Z Z sin t cos(2t) dt = sin t(2 cos2 t − 1) dt Z Z Z 2 = 2 sin t cos t dt − sin t dt = 2 sin t cos2 t dt + cos t On the remaining integral, using direct substitution with u = cos t and du = − sin t dt, we have that: Z Z 2 2 2 2 sin t cos t dt = −2u2 du = − u3 + C = − cos3 t + C 3 3

Page 3 of 22

MATH 105 921

Solutions to Integration Exercises

Therefore, Z

7)

Z

2 sin t cos(2t) dt = − cos3 t + cos t + C 3

x+1 dx 4 + x2 Solution: Observe that we may split the integral as follows: Z Z Z x+1 x 1 dx = dx + dx 4 + x2 4 + x2 4 + x2 On the first integral on the right hand side, we use direct substitution with u = 4+x2 , and du = 2x dx. We get: Z Z x 1 dx = du = ln |2u| + C = ln(8 + 2x2 ) + C 2 4+x 2u On the second integral on the right hand side, we use inverse trigonometric substitux ), so 2 sec2 t dt = dx. Thus, tion with 2 tan t = x (or equivalently, t = arctan 2 Z Z Z 1 1 2 sec2 t 2 dx = 2 sec t dt = dt 4 + x2 4 + 4 tan2 t 4 sec2 t Z x 1 t 1 +C = dt = + C = arctan 2 2 2 2 Therefore, Z Z Z x+1 x 1 dx = dx + dx 4 + x2 4 + x2 4 + x2

8)

Z

= ln(8 + 2x2 ) +

x 1 +C arctan 2 2

sin(tan θ) dθ cos2 θ Solution: Using direct substitution with u = tan θ and du = sec2 θ dθ, we get: Z Z Z sin(tan θ) 2 dθ = sec θ sin(tan θ) dθ = sin u du = − cos u + C cos2 θ Z sin(tan θ) ⇒ dθ = − cos(tan θ) + C cos2 θ

Page 4 of 22

MATH 105 921 9)

Solutions to Integration Exercises

√ x 3 − 2x − x2 dx

Z

Solution: Completing the square, we get 3 − 2x − x2 = 4 − (x + 1)2 . Using direct substitution with u = x + 1 and du = dx, we get: Z √ Z Z √ Z √ √ 2 2 2 4 − u2 du x 3 − 2x − x dx = (u − 1) 4 − u du = u 4 − u du − For the first integral on the right hand side, using direct substitution with t = 4 − u2 , and dt = −2u du, we get: Z √ Z 3 1 1√ 1 3 2 u 4 − u du = − t dt = − t 2 + C = − (4 − u2 ) 2 + C 2 3 3 For the second integral on the right hand
side, using inverse trigonometric substitution u with 2 sin s = u, that is, s = arcsin 2 , and 2 cos s ds = du, we get: Z p Z Z √ 2 2 4 − u du = 4 − 4 sin s2 cos s ds = 4 cos2 s ds Z 1 + cos(2s) ) = (2 + 2 cos(2s)) ds (using half-angle formula cos2 s = 2 = 2s + sin(2s) + C = 2s + 2 sin s cos s + C (using double-angle formula sin(2s) = 2 sin s cos s) u u u = 2 arcsin + 2 sin(arcsin ) cos(arcsin )+C 2 2 2 √   u 4 − u2 +u = 2 arcsin +C 2 2 Therefore, Z √ Z √ Z √ 4 − u2 du x 3 − 2x − x2 dx = u 4 − u2 du −

√  u 3 4 − u2 1 2 2 = − (4 − u ) − 2 arcsin −u +C 3 2 2 ! p   Z √ 2 4 − (x + 1) 3 x + 1 1 ⇒ x 3 − 2x − x2 dx = − (4 − (x + 1)2 ) 2 − 2 arcsin − (x + 1) +C 3 2 2

10)

Z

π 3

sin3 z cos z dz

0

Page 5 of 22

MATH 105 921

Solutions to Integration Exercises

Solution: Using direct substitution √ with u = sin z, and du = cos z dz, when z = 0, π then u = 0, and when z = 3 , u = 23 . We have that: Z ⇒

11)

Z

3x2

Z

π 3

3

sin z cos z dz = 0 π 3

sin3 z cos z dz =

0

Z

√

3 2

0

u4 23 9 |0 = u du = 4 64 √

3

9 64

1 dx + 2x + 1

2 1 2 Solution: Completing the square, we get that 3x + 2x + 1 = 3 x + + = 3 3 !   2  1 1 3 2 9 x+ , and du = + 1 . Using direct substitution with u = √ x + 3 2 3 3 2 3 √ dx, we get: 2 Z Z Z 1 1 1 3 √ du = √ arctan u + C dx = dx = 1 2 9 2 3x + 2x + 1 2( 2 (x + 3 ) + 1) 2(u2 + 1) 2    Z 1 1 1 3 +C dx = √ arctan √ x + ⇒ 2 3x + 2x + 1 3 2 2 2

12)

Z

et



1 dt +1

1 Solution: Using direct substitution with u = et + 1 and du = et dt, so dt = t du = e 1 du. Hence, we get: u−1 Z Z 1 1 dt = du t e +1 u(u − 1) Using partial fraction, we get: 1 A B A(u − 1) + Bu (A + B)s + (−A) = + = = u(u − 1) u u−1 u(u − 1) u(u − 1)

Page 6 of 22

MATH 105 921

Solutions to Integration Exercises

Thus, A + B = 0 and −A = 1. So, A = −1, and B = 1. Thus, we have that: Z Z Z −1 1 1 du = du + du u(u − 1) u u−1 Therefore, Z

Z Z −1 1 1 du = du + du = − ln |u| + ln |u − 1| + C u(u − 1) u u−1 Z 1 ⇒ dt = − ln |et + 1| + ln |et | + C = − ln |et + 1| + t + C t e +1

13)

Z

e3a cos(3a) da

Solution: Using direct substitution with t = 3a, and dt = 3 da, we get: Z Z 1 t 3a e cos(3a) da = e cos t dt 3 Using integration by parts with u = cos t, du = − sin t dt, and dv = et dt, v = et , we get: Z Z 1 t 1 1 t e cos t dt = e cos t + et sin t dt 3 3 3 Using integration by parts again on the remaining integral with u1 = sin t, du1 = cos t dt, and dv1 = et dt, v1 = et , we get: Z Z 1 1 1 t t e sin t dt = sin te − et cos t dt 3 3 3 Thus, 1 t e cos t dt = 3 Z 1 t e cos t dt = ⇒ 3 Z

1 t e cos t + 3 1 t e cos t + 6

1 1 sin tet − 3 3 1 t e sin t + C 6

Z

et cos t dt

Therefore, Z

1 1 e3a cos(3a) da = e3a cos(3a) + e3a sin(3a) + C 6 6

Page 7 of 22

MATH 105 921 14)

Z

Solutions to Integration Exercises

x2 dx 1 + x6 Solution: Using direct substitution with u = x3 , and du = 3x2 dx, we get: Z Z 1 1 1 x2 dx = du = arctan u + C = arctan(x3 ) + C 1 + x6 3(1 + u2 ) 3 3

15)

Z

1 dt t(ln t)2 Solution: Using direct substitution with u = ln(t) and du = 1t dt, we get: Z Z 1 1 1 dt = du = − +C t(ln t)2 u2 u Z 1 1 dt = − +C ⇒ t(ln t)2 ln t

16)

Z

xe2x dx (2x + 1)2 Solution: Using integration by parts with u = xe2x , du = (e2x + 2xe2x ) dx, and 1 dv = (2x + 1)−2 dx, v = − 2(2x+1) , we get: Z

xe2x xe2x dx = − + (2x + 1)2 2(2x + 1)

Z

e2x + 2xe2x dx 2(2x + 1)

On the remaining integral, using direct substitution with u = 2x + 1, and du = 2 dx, we get: Z u−1 Z Z 2x e + (u − 1)eu−1 1 u−1 1 1 e + 2xe2x dx = du = e du = eu−1 + C = e2x + C 2(2x + 1) 4u 4 4 4 Therefore, Z

xe2x xe2x 1 2x e2x dx = − + e + C = +C (2x + 1)2 2(2x + 1) 4 4(2x + 1)

Page 8 of 22

MATH 105 921 17)

Z

Solutions to Integration Exercises

(tan x + cot x)2 dx Solution: Z Z 2 (tan x + cot x) dx = (tan2 x + 2 tan x cot x + cot2 x) dx Z = (sec2 x − 1 + 2 + csc2 x − 1) dx (using identities for tan2 x and cot2 x) Z = (sec2 x + csc2 x) dx = tan x − cot x + C

18)

Z

2

tet sin(t2 ) dt Solution: Using direct substitution with x = t2 and dx = 2t dt, we get: Z Z 1 t2 2 te sin(t ) dt = ex sin x dx 2 Using integration by parts with u = sin x, du = cos x dx, and dv = ex dx, v = ex , we get: Z Z 1 x 1 1 x e sin x dx = e sin x − ex cos x dx 2 2 2 Using integration by parts again on the remaining integral with u1 = cos x, du1 = − sin x dx, and dv1 = ex dx, v1 = ex , we get: Z Z 1 1 x 1 x e cos x dx = e cos x + ex sin x dx 2 2 2 Thus, 1 x e sin x dx = 2 Z 1 x ⇒ e sin x dx = 2 Z

1 x e sin x − 2 1 x e sin x − 4

1 x 1 e cos x − 2 2 1 x e cos x + C 4

Z

ex sin x dx

Therefore, Z

1 2 1 2 2 tet sin(t2 ) dt = et sin(t2 ) − et cos(t2 ) + C 4 4

Page 9 of 22

MATH 105 921 19)

Solutions to Integration Exercises

2p − 4 dp p2 − p

Z

Solution: Using partial fraction, we get: A B A(p − 1) + Bp (A + B)p + (−A) 2p − 4 = + = = p(p − 1) p p−1 p(p − 1) p(p − 1) Thus, A + B = 2 and −A = −4. So, A = 4, and B = −2. We have that: Z Z Z 2p − 4 4 2 dp = dp − dp p(p − 1) p p−1 Z 2p − 4 dp = 4 ln |p| − 2 ln |p − 1| + C ⇒ p(p − 1)

20)

Z

4 3

1 dx (3x − 7)2

Solution: Using direct substitution with u = 3x − 7, and du = 3 dx, when x = 3, then u = 2, and when x = 4, u = 5. We have that: Z 5 Z 4 1 −1 5 1 1 1 1 dx = du = |2 = − + = 2 2 3u 15 6 10 2 3u 3 (3x − 7) Z 4 1 1 ⇒ dx = 2 10 3 (3x − 7)

21)

Z

t3 5

(2 − t2 ) 2

dt

Solution: Using direct substitution with u = 2 − t2 , and du = −2t dt, we get: Z Z Z 2−u t2 t3 − 5 dt = 5 (t dt) = 5 du (2 − t2 ) 2 (2 − t2 ) 2 2u 2 Z 5 1 3 = (−u− 2 + u− 2 ) du 2 1 2 3 = u− 2 − u− 2 + C 3 Z 1 3 2 t3 (2 − t2 )− 2 − (2 − t2 )− 2 + C ⇒ 5 dt = 3 (2 − t2 ) 2

Page 10 of 22

MATH 105 921 22)

Z

x2

√

Solutions to Integration Exercises

1 dx 4 − x2

Solution:
Using inverse trigonometric substitution with x = 2 sin y, that is, y = x arcsin 2 , and dx = 2 cos y dy, we get: Z Z Z 1 2 cos y 2 cos y √ p dx = dy = dy 4 sin2 y(2 cos y) x2 4 − x2 4 sin2 y 4 − 4 sin2 y Z 1 1 csc2 y dy = − cot y + C = 4 4 Therefore, Z

√ x 4 − x2 1 1 √ )+C =− +C dx = − cot(arcsin 4 2 4x x2 4 − x2

Z p 23) y 2 − 1 dy Solution:   Using inverse trigonometric substitution with y = sec u, that is, u = arccos y1 , and dy = sec u tan u du, we get: Z p Z √ Z 2 2 y − 1 dy = sec u − 1(sec u tan u du) = tan2 u sec u du Z Z Z 2 3 = (sec u − 1) sec u du = sec u du − sec u du

For the second integral on the right hand side, we have that: Z sec u du = ln | sec u + tan u| + C For the first integral on the right hand side, we use the reduction formula: Z Z 1 1 1 1 3 sec u du = tan u sec u + ln | sec u + tan u| + C sec u du = tan u sec u + 2 2 2 2   p Observe that since u = arccos y1 , we have that tan u = y 2 − 1. Therefore, Z

1 1 sec u du − sec u du = tan u sec u − ln | sec u + tan u| + C 2 2 Z p p p 1 1 y 2 − 1 dy = y y 2 − 1 − ln |y + y 2 − 1| + C ⇒ 2 2 3

Z

Page 11 of 22

MATH 105 921 24)

Z

Solutions to Integration Exercises

x sin x cos x dx

Solution: Using the double angle identity sin(2x) = 2 sin x cos x, we have that: Z Z 1 x sin x cos x dx = x sin(2x) dx 2 Using direct substitution with t = 2x, and dt = 2 dx, we get: Z Z 1 1 x sin(2x) dx = t sin t dt 2 8 Using integration by parts with u = t, du = dt, and dv = sin t dt, v = − cos t, we get: Z Z 1 1 1 1 1 t sin t dt = − t cos t + cos t dt = − t cos t + sin t + C 8 8 8 8 8 Therefore, Z

25)

Z

1 1 x sin x cos x dx = − x cos(2x) + sin(2x) + C 4 8

(1 + cos θ)2 dθ

Solution: Z Z 2 (1 + cos θ) dθ = (1 + 2 cos θ + cos2 θ) dθ Z Z Z = dθ + 2 cos θ dθ + cos2 θ dθ  Z  1 + cos(2θ) dθ = θ + 2 sin θ + 2 θ sin(2θ) = θ + 2 sin θ + + +C 2 4 Z 1 3 ⇒ (1 + cos θ)2 dθ = θ + 2 sin θ + sin(2θ) + C 2 4

26)

Z

√

1 dx 4x − x2

Page 12 of 22

(using half-angle formula)

MATH 105 921

Solutions to Integration Exercises

Solution: Completing the square yields 4x − x2 = 4 − (x − 2)2 . Using direct substitution with u = x − 2, and du = dx, we get: Z Z 1 1 √ √ dx = du 4x − x2 4 − u2 u Using inverse trigonometric substitution with u = 2 sin t, that is, t = arcsin , and 2 du = 2 cos t dt, we get: Z Z Z Z 2 cos t 2 cos t 1 √ p dt = dt = t + C du = dt = 2 cos t 4 − u2 4 − sin2 t   Z 1 x−2 √ ⇒ +C dx = arcsin 2 4x − x2 27)

Z

1

1 1

0

1 + x3

dx

1 −2 x 3 dx, so dx = 3 2 3x 3 du = 3(u − 1)2 du. When x = 0, u = 1 and when x = 1, u = 2. We have that: Z 2 Z 2 Z 1 3 1 3(u − 1)2 (3u − 6 + ) du du = 1 dx = u u 1 1 0 1 + x3 3 = ( u2 − 6u + 3 ln |u|) |21 2 3 3 = (6 − 12 + 3 ln 2) − ( − 6 + 3 ln 1) = − + 3 ln 2 2 2 Z 1 3 1 + 3 ln 2. ⇒ 1 dx = − 2 0 1 + x3 1

Solution: Using direct substitution with u = 1 + x 3 , and du =

28)

Z

1 dx x3 + x Solution: Using partial fractions, we have: 1 A Bx + C A(x2 + 1) + (Bx + C)x (A + B)x2 + Cx + A = + = = x3 + x x x2 + 1 x3 + x x3 + x So, A + B = 0, C = 0 and A = 1. So, B = −1 and we get: Z Z Z Z 1 x x 1 dx = dx − dx = ln |x| − dx 3 2 2 x +x x x +1 x +1

Page 13 of 22

MATH 105 921

Solutions to Integration Exercises

On the remaining integral, using direct substitution with u = x2 + 1 and du = 2x dx, we get: Z Z x 1 1 1 dx = du = ln |u| + C = ln(x2 + 1) + C 2 x +1 2u 2 2 Therefore, Z

x3

1 1 dx = ln |x| − ln(x2 + 1) + C +x 2

Remark: This involves partial fractions with non-linear factors, which you are not required to master in this course! 29)

Z

ln(1 + t) dt Solution: Using direct substitution with s = 1 + t, and ds = dt, we have that: Z Z ln(1 + t) dt = ln s ds 1 ds, and dv = ds, v = s, we get: s Z Z Z 1 ds = s ln s − s + C ln s ds = s ln s − s ds = s ln s − s

Using integration by parts with u = ln s, du =

Therefore, Z

30)

Z

ln(1 + t) dt = (1 + t) ln(1 + t) − (1 + t) + C

sin(3x) cos(5x) dx Solution: Using the trigonometric identity that sin a cos b = 12 (sin(a+b)+sin(a−b)), we get: Z Z 1 1 1 (sin(8x) + sin(−2x)) dx = − cos(8x) + cos(−2x) + C sin(3x) cos(5x) dx = 2 16 4 Remark: You are not required to memorize any sum to product or product to sum trigonometric identities!

Page 14 of 22

MATH 105 921 31)

Z

k2

Solutions to Integration Exercises

1 dk − 6k + 9

Solution: By completing the square, we observe that k 2 − 6k + 9 = (k − 3)2 . So, using direct substitution with u = k − 3, and du = dk, we have that: Z Z Z 1 1 1 1 dk = du = − dk = +C k 2 − 6k + 9 (k − 3)2 u2 u Z 1 1 dk = − +C ⇒ 2 k − 6k + 9 k−3 32)

Z

1 dx sec x − 1 1 , we get: cos x  Z Z Z  Z 1 1 cos x 1 −1 + dx = −x + dx = dx = dx sec x − 1 1 − cos x 1 − cos x 1 − cos x x , so dt = For the remaining integral, use a direct substitution with t = tan 2       √ x 1 1 x x dx. We also can compute that sec = t2 + 1, cos = √ sec2 2 2 2 2 t2 + 1 x t 2 dt. Using double angle formula, we get: =√ and sin . So, dx = 2 2 t +1 t2 + 1

Solution: Since sec x =

2

cos x = cos

x 2

− sin

2

x 2

=

1 t2 1 − t2 − = t2 + 1 t2 + 1 t2 + 1

So, after the substitution, we get:   Z Z Z 2 1 1 1 dt = dt dx = 2 2 1−t 1 − cos x t2 1 − t2 +1 t + 1 x 1 = − + C = − cot +C t 2

Therefore,

Z

x 1 +C dx = −x − cot sec x − 1 2

Remark: This is an extremely challenging question; do not panic if you do not know how to solve it!

Page 15 of 22

MATH 105 921 33)

Z

1

Solutions to Integration Exercises

2 dx +1

e−x

0

Solution: Using direct substitution with u = e−x + 1, and du = −e−x dx, that is 1 dx = − u−1 du. When x = 0, u = 2, and when x = 1, u = e−1 + 1. So, we get: Z

1 0

2 dx = e−x + 1

Z

e−1 +1 2

−2 du u(u − 1)

Using partial fraction, we get: A B A(u − 1) + Bu (A + B)s + (−A) −2 = + = = u(u − 1) u u−1 u(u − 1) u(u − 1) Thus, A + B = 0 and −A = −2. So, A = 2, and B = −2. Thus, we have that: Z

e−1 +1 2

−2 du = u(u − 1)

Z

e−1 +1 2

2 du − u

Z

e−1 +1 2

2 du u−1

Therefore, Z

34)

Z

e−1 +1 2

c2

−2 −1 du = (2 ln |u| − 2 ln |u − 1|) |e2 +1 = (2 ln(e−1 + 1) + 2) − (2 ln 2 − 0) u(u − 1) Z 2 ⇒ dx = 2 ln(e−1 + 1) + 2 − 2 ln 2. −x e +1 1 dc − 6c + 10

Solution: Completing the square yields c2 − 6c + 10 = (c − 3)2 + 1. So, using direct substitution with u = c − 3, and du = dc, we have that: Z Z Z 1 1 1 dc = dc = du = arctan u + C 2 2 2 c − 6c + 10 (c − 3) + 1 u +1 Z 1 ⇒ dc = arctan(c − 3) + C 2 c − 6c + 10

35)

Z

f (x)f ′ (x) dx

Page 16 of 22

MATH 105 921

Solutions to Integration Exercises

Solution: Using direct substitution with u = f (x), and du = f ′ (x) dx, we get: Z Z 1 ′ f (x)f (x) dx = u du = u2 + C 2 Z 1 ⇒ f (x)f ′ (x) dx = (f (x))2 + C 2

36)

Z

x2

1 dx + 4x + 5

Solution: Completing the square, we get x2 + 4x + 5 = (x + 2)2 + 1. Using direct substitution with u = x + 2 and du = dx, we get: Z Z Z 1 1 1 dx = dx = du = arctan(u) + C 2 2 2 x + 4x + 5 (x + 2) + 1 u +1 Z 1 ⇒ dx = arctan(x + 2) + C x2 + 4x + 5

37)

Z

2 0

1 dx (3 + 5x)2

Solution: Using direct substitution with u = 3 + 5x, and du = 5 dx, when x = 0, then u = 3, and when x = 2, u =...