www.ck12.org. Chapter 4. Integration, Solution Key. 4.7 Integration by Substitution. 1. Let u = ln x and dv = x. Then du = 1 x dx and v = x2. 2 . ∫ x ln x dx = uv−lim ...
4.7 Integration by Substitution 1. Let u = ln x and dv = x. Then du = 1x dx and v =
Z
x2 2.
x ln x dx = uv − lim v du x2 ln x x2 1 − · dx 2 2 x Z x x2 ln x − dx = 2 2 x2 ln x x2 = − +C 2 4 Z
=
2. Let u = ln x and dv =
√ 3 x. Then du = 1x dx and v = 23 x 2 .
Z Z3
x ln x dx = uv −
Z
v du 3
3
Z √ 3 1 2x 2 ln x 2 xln x dx = − x 2 · dx 3 3 x
1
1 3 2
2x ln x 2 = − 3 3
Z3
1
x 2 dx 1
=
3 2
3
2x ln x 4x 2 − 3 9
! 3 1
3 2
3 2
3
3
4 2 × 3 ln x 4 × 3 − − 0− = 3 9 9 2 × 3 2 ln x 4 × 3 2 4 = − + 3 9 9 3. Let u2 = 2x + 1. Then
u2 = 2x + 1 u2 − 1 = 2x u2 − 1 =x 2 u du = dx 157
4.7. Integration by Substitution
www.ck12.org
Z
1 u2 − 1 u du u 2 Z 2 u −1 = du 2 Z 1 u2 − 1 du = 2 1 u2 = −u 2 3 " √ # 3 √ 2x + 1 1 = − 2x + 1 2 3 1√ 2x + 1 = 2x + 1 − 1 +C 2 3 √ 2x + 1 (2x − 2) 1 = · +C 2√ 3 2x + 1 (x − 1) = +C 3
x √ dx = 2x + 1
Z
4. Let u = 1 − x2 . Then
du = −2x dx 1 − du = x dx 2 and u − 1 = −x2 − u + 1 = x2
When using u−substitution, just put limits as u1 and u2 as placeholders on the integral. After u is replaced by the function of x, put back the original limits of integration. 158
www.ck12.org
Chapter 4. Integration, Solution Key
Z1
Zu2 p p x 1 − x2 dx = x2 x 1 − x2 dx 3
u1
0
1 = 2 =
1 2
1 = 2
Zu2
− (−u + 1)
√ u du
u1 Zu2
1 3 u 2 − u 2 du
u1 5
u2 5 2
3
−
u2
!
3 2
1 3 5 2 1 2 1 − x2 2 − 1 − x2 2 = 2 5 3 0 1 2 2 − = 0− 2 5 3 2 = 15 5. Let u = x and dv = cos x dx. Then du = dx and v = sin x. Z
x cos x dx = x sin x −
Z
sin x dx
= x sin x − (− cos x) +C = x sin x + cos x +C 6. Let u = x3 + 9. Then
3 x-8. +C. 2. Let u = 2+x. Then du = dx. â« /. 2+xdx = â«. (2+x). -1. 2 dx. = â« u. 1. 2 du. = u. 3. 2. 3. 2. +C ... Integration Techniques, Solution Key. 5. Let u = e-x +2.
sec3 x dx = â« sec2 xsecx dx. Let u = sec x and dv = sec2 x dx. Then du = tan x dx and v = tanx. â« sec3 x dx = â« sec2 xsecx dx. = uv-. â« v du. = (secx)(tanx)-. â«.
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Integration by Partial Fractions www.ck12.org. 7.3 Integration by Partial Fractions. 1. 1 x2 -1. = A x+1. +. B x-1. 1 = A(x-1)+B(x+1). Let x = 1. 1 = A(1-1)+B(x+1).
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