(1.) ∫[x/(x - 1)4]dx here is the problem
Let u = x - 1
du = dx
x = u + 1
∫(u + 1)du/u4
make substitutions
= ∫(u-3 + u-4)du divide thru by u4
= (-1/2)u-2 - (1/3)u-3
+ C integrate
= (-1/2)(x - 1)-2 - (1/3)(x -
1)-3 + C replace u with x - 1
-1 1
= ____________ - __________ + C use positive exponents
2(x - 1)2 3(x - 1)3
(2.) ∫ dx/(4 + 5x1/2) here is the problem
let u = 5x1/2
u2 = 25x square each side
2u du = 25 dx take the derivative of each side
(2/25)u du = dx multiply by 1/25 each side
(2/25)∫(u du)/(4 + u) make
substitutions
= (2/25)∫(u + 4 - 4)du/(4 + u)
add 4 and subtract 4
= (2/25)∫du - (8/25)∫[du/(4 +
u)] separate like this
= (2/25)u - (8/25) ln (4 + u) + C integrate
= (2/25)(5x1/2) - (8/25) ln
(4 + 5x1/2) + C replace u
with 5x1/2
(3.) ∫x(1 + 2x)2/3 dx
Let u = 1 + 2x
du = 2 dx
(1/2)du = dx
(1/2)u - (1/2) = x
∫[(1/2)u - (1/2)](u2/3)[(1/2)du] make substitutions
= (1/4)∫u5/3 du - (1/4)∫u2/3 du
multiply thru
= (1/4)(3/8)u8/3 -
(1/4)(3/5)u5/3 + C integrate
= (3/32)u8/3 - (3/20)u5/3
+ C multiply
= (3/32)(1 + 2x)8/3 -
(3/20)(1 + 2x)5/3 + C make
substitutions
(4.) ∫(3x5)/(1 + x3)3/2 dx
here is the problem
Let u = 1 + x3
x3 = u - 1
x = (u - 1)1/3
dx = (1/3)(u - 1)-2/3
du
3∫[(u - 1)5/3 * (1/3)(u - 1)-2/3]/u3/2 make substitutions
= ∫(u - 1)/u3/2
du simplify, add exponents
= ∫[u-1/2 - u-3/2]du divide
= 2u1/2 + 2u-1/2 +
C integrate
= 2(1 + x3)1/2 +
2(1 + x3)-1/2 + C
make substitutions
(5.) ∫(x8)(1 - x3)1/3 dx here is the problem
Let u = 1 - x3
u - 1 = -x3
1 - u = x3
(1 - u)1/3 = x
dx = -(1/3)(1 - u)-2/3 du
∫(1 - u)8/3 * u1/3 * (-1/3)(1 - u)-2/3
du make substitutions
= (-1/3)∫(1 - u)2 * u1/3 du add exponents
= (-1/3)∫(1 - 2u + u2)(u1/3)du square the binomial
= (-1/3)∫(u1/3 - 2u4/3 + u7/3)du multiply thru parentheses
= (-1/3)[(3/4)u4/3 - (6/7)u7/3
+ (3/10)u10/3 + C]
integrate
= (-1/4)u4/3 + (2/7)u7/3
- (1/10)u10/3 + C multiply
thru
= (-1/4)(1 - x3)4/3
+ (2/7)(1 - x3)7/3 - (1/10)(1 - x3)10/3
+ C
[make substitutions]
(6.) ∫(x8)(1 - x3)6/5 dx here is the problem
Let u = 1 - x3
x3 = 1 - u
x = (1 - u)1/3
dx = (-1/3)(1 - u)-2/3
du
∫(1 - u)8/3 * u6/5 * (-1/3)(1 - u)-2/3
du make substitutions
= (-1/3)∫(1 - u)2 * u6/5 du add exponents
= (-1/3)∫(1 - 2u + u2)(u6/5)du square the binomial
= (-1/3)∫[u6/5 - 2u11/5 + u16/5]du multiply thru parentheses
= (-1/3)[(5/11)u11/5 - (5/8)u16/5 + (5/21)u21/5]
+ C integrate
= (-5/33)u11/5 + (1/24)u16/5
- (5/63)u21/5 + C multiply
thru
= (-5/33)(1 - x3)11/5
+ (1/24)(1 - x3)16/5 - (5/63)(1 - x3)21/5
+ C
[make substitutions]
(7.) ∫x(1 + x)1/2 dx here is the problem
Let u = 1 + x
x = u - 1
dx = du
∫(u - 1)(u)1/2 du make substitutions
= ∫[u3/2 - u1/2]du multiply thru parentheses
= (2/5)u5/2 - (2/3)u3/2
+ C integrate
= (2/5((1 + x)5/2 - (2/3)(1 +
x)3/2 + C make substitutions
(8.) ∫ (1 + x)/(1 - x)1/2 dx
here is the problem
Let u = 1 - x
du = -dx
-du = dx
x = 1 - u
-∫(2 - u)/(u1/2) du make substitutions
= ∫(u - 2)/(u1/2)
du simplify
= ∫[u1/2 - 2u-1/2]du divide
= (2/3)u3/2 - 4u1/2
+ C integrate
= (2/3)(1 - x)3/2 - 4(1 - x)1/2
+ C make substitutions
(9.) ∫ (3x2 - x)/(1
+ x)1/2 dx
Let u = 1 + x
du = dx
x = u - 1
∫[3(u - 1)2 - (u - 1)]du/(u1/2) make substitutions
= ∫[(3u2 - 6u + 3) - (u - 1)]du/(u1/2) square the binomial
= ∫(2u2 - 7u + 4)du/(u1/2) combine like terms on top
= ∫[2u3/2 - 7u1/2 + 4u-1/2]du divide
= (4/5)u5/2 - (14/3)u3/2
+ 8u1/2 + C integrate
= (4/5)(1 + x)5/2 - (14/3)(1
+ x)3/2 + 8(1 + x)1/2 + C
[make substitutions]
(10.) ∫ (x3 - x)/(x2
- 2)1/2 dx
Let u = x2 - 2
x2 = u + 2
x = (u + 2)1/2
dx = (1/2)(u + 2)-1/2
du
(1/2)∫[(u + 2)3/2 - (u + 2)1/2](u + 2)-1/2 du make substitutions
= (1/2)∫[(u + 2) - 1]du
multiply thru add exponents
= (1/2)∫(u + 1)du
combine like terms
= (1/2)(1/2)u2 + (1/2)u +
C integrate
= (1/4)(x2 - 2)2 + (1/2)(x2 - 2) + C
[make substitutions & multiply]