Page 116 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 5] INTEGRALS 107

n
(a)Use integration by parts, letting u ¼ ln x, dv ¼ x dx,sothat du ¼ðdxÞ=x, v ¼ x nþ1 =ðn þ 1Þ. Then
ð ð ð x nþ1 ð x nþ1 dx
n
x ln xdx ¼ udv ¼ uv vdu ¼ ln x
n þ 1 n þ 1 x
x nþ1 x nþ1
þ c
¼ ln x 2
n þ 1
ðn þ 1Þ
ð ð 1
2
x 1 ln xdx ¼ ln xdðln xÞ¼ ðln xÞ þ c:
2
ðbÞ
ð
ffiffiffiffiffiffiffiffi
p 2xþ1
5.18. Find 3 dx.
ð
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 y
Let 2x þ 1 ¼ y,2x þ 1 ¼ y . Then dx ¼ ydy and the integral becomes 3 ydy.
y
y
Integrate by parts, letting u ¼ y, dv ¼ 3 dy;then du ¼ dy, v ¼ 3 =ðln 3Þ, and we have
ð ð ð y 3 y ð 3 y y 3 y 3 y
y
ln 3 ln 3 ln 3
3 ydy ¼ udv ¼ uv vdu ¼ dy ¼ 2 þ c
ðln 3Þ
1
ð
5.19. Find x lnðx þ 3Þ dx.
0
dx x 2
. Hence on integrating by parts,
x þ 3 2
Let u ¼ lnðx þ 3Þ, dv ¼ xdx. Then du ¼ , v ¼
ð 2 ð 2 2 ð
x 1 x dx x 1 9
dx
x lnðx þ 3Þ dx ¼ lnðx þ 3Þ ¼ lnðx þ 3Þ x 3 þ
2 2 x þ 3 2 2 x þ 3
( )
x 2 1 x 2
3x þ 9lnðx þ 3Þ þ c
2 2 2
¼ lnðx þ 3Þ
1 5 9
ð
Then x lnðx þ 3Þ dx ¼ 4ln 4 þ ln 3
0 4 2
ð
6 x
5.20. Determine dx.
ðx 3Þð2x þ 5Þ
6 x A B
Use the method of partial fractions. Let ¼ þ .
x 3 2x þ 5
ðx 3Þð2x þ 5Þ
Method 1: To determine the constants A and B, multiply both sides by ðx 3Þð2x þ 5Þ to obtain
or 6 x ¼ 5A 3B þð2A þ BÞx
6 x ¼ Að2x þ 5Þþ Bðx 3Þ ð1Þ
Since this is an identity, 5A 3B ¼ 6, 2A þ B ¼ 1 and A ¼ 3=11, B ¼ 17=11. Then
6 x 3=11 17=11 3 17
ð ð ð
ln j2x þ 5jþ c
x 3 2x þ 5 11 22
dx ¼ dx þ dx ¼ ln jx 3j
ðx 3Þð2x þ 5Þ
Method 2: Substitute suitable values for x in the identity (1). For example, letting x ¼ 3 and x ¼ 5=2in
(1), we find at once A ¼ 3=11, B ¼ 17=11.
dx
ð
5.21. Evaluate by using the substitution tan x=2 ¼ u. 2
5 þ 3 cos x √1 + u u
From Fig. 5-7 we see that x/2
u 1 1
ffiffiffiffiffiffiffiffiffiffiffiffiffi ;
sin x=2 ¼ p cos x=2 ¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ u 2 1 þ u 2 Fig. 5-7