144
CHAPTER 19
19.24
Solve Problem 19.23 by using Problem 19.1.
Notice that D,(4 - 2/
2 ) = -4t. Then
19.25
Find
Then
19.26 Evaluate
Then
19.27
Find
Note that x
2 -2x + 1 = (x - I)
2
.
Then
19.28
Find J (A4 + 1)"V dx.
Let u = x
4 + l, du=4x*dx Note that
Hence,
19.29 Evaluate
Let u = 1 + 5x
2 , du = 10* dx. Then
19.30
Find J xVax + b dx, when a^O.
Then
Let u = ax + b, du = adx. Note that x = (u-b)/a.
19.31 Evaluate
Let u = sin 3x. By the chain rule, du = 7> cos 3x dx. So,
19.32 Find J VF
7
* x
2 dx.
So, let H = X — 1, du = dx.
Let
w = x
3 + 5, rfw = 3x
2 dx.
Let u = x + l, du — dx.
I Let U = !-A:, du = -dx. Note that x = 1 - u. Then J VI - x x
2 dx = J Vw(l - w)
2 • (-1) du =
-JVI7(l-2 M + M
2 )rf M = -;(«"
2 -2 M
3 '
2 + M
5 'V M = -[t M
3 '
2 -2(i)«
5/2 +| M
7 '
2 ]+C= -2 M
3/2 (i-| M +
|«
2 ) + C = -2(VT^)
3
[i - 1(1 - x) + J(l - ^:)
2
] + C = - Tfe(VT^)
3
(8 + 12x + 15x
2
) + C.