HW 7 Solution
(1)
Z
(7.4 Problem 22)
3x2 + 4x + 3
dx
(x2 + 1)2
2
= Ax+B
+ (xCx+D
Solution: From partial fraction 3x(x2+4x+3
2 +1)2 ,
+1)2
x2 +1
we have 3x2 + 4x + 3 = (Ax + B)(x2 + 1) + (CX + D), 3x2 + 4x + 3 =
Ax3 +AX +Bx2 +B +CX +D and 3x2 +4x+3 = Ax3 +Bx2 +(A+C)X +B +D
which gives A = 0, B = 3, A + C = 4 and B + D = 3. ThusRwe get A = 0
2
2
B = 3, C = 4 and D = 0 and 3x(x2+4x+3
= x23+1 + (x24x
. Thus 3x(x2+4x+3
dx =
+1)2
+1)2
+1)2
R 3
R 4x
dx + (x2 +1)2 dx = 3 arctan(x) − x22+1 + C. We use the substitution
x2 +1
R
R
dx = 2u
dx = − u2 +C = − x22+1 +C.
u = x2 +1 and du = 2xdx to find (x24x
+1)2
u2
(2)
Z
x2
4x + 3
dx
+ 2x + 10
Solution: Completing the square, we get x2 + 2x + 10R = (x + 1)2 +
4x+3
33 . Let x + 1 = 3u. Then dx = 3du and x = 3u − 1.
2 +2x+10 dx =
x
R 4x+3
R 4·(3u−1)+3
R 12u−1
R 12u−1
R 4u
3du = 9(u
du −
2 +33 dx =
2 u+33
2 +1) 3du =
2 +1) du =
(x+1)
3
3(u
u2 +1
R 1
1
1
x+1 2
1
x+1
2
du = 2 ln |u +1|− 3 arctan(u)+C = 2 ln |( 3 ) +1|− 3 arctan( 3 )+C.
3
u2 +1
(3)
Z
x3 − 2x2 − 2x + 3
dx
x2 (x2 + 1)
3
2
−2x+3
Solution: From the partial fraction x −2x
= Ax + xB2 + Cx+D
, we get
x2 (x2 +1)
x2 +1
3
2
2
2
2
3
2
x −2x −2x+3 = Ax(x +1)+B(x +1)+(Cx+D)x , x −2x −2x+3 = Ax3 +
Ax+Bx2 +B+Cx3 +Dx2 and x3 −2x2 −2x+3 = (A+C)x3 +(B+D)x2 +Ax+B.
Thus A + C = 1, B + D = −2, A = −2 and B = 3 HenceA = −2,
3
2 −2x+3
= − x2 + x32 + 3x−5
and
B = 3, C = 3 and D = −5. Thus x −2x
x2 (x2 +1)
x2 +1
R x3 −2x2 −2x+3
3
3
dx = −2 ln |x| − x + 2 ln |x2 + 1| − 5 arctan(x) + C
x2 (x2 +1)
(4) (Problems from Sec 7.4) Determine whether each integral is convergentRor divergent. If the integral is convergent, compute its value.
∞
(a) 1 12 dx
x3
R∞ 1
Rb 1
1 b
1
3
Solution: 1 2 dx = limb→∞ 1 2 dx = limb→∞ 3x == limb→∞ 3b 3 −
3
3
x
x
1
R∞
3 = ∞. So 1 12 dx diverges.
x3
R∞
(b) e x ln1 x dx
b
R∞
Rb
Solution: e x ln1 x dx = limb→∞ 1 x ln1 x dx = limb→∞ ln | ln x| == limb→∞ ln | ln b|−
e
R∞
ln | ln e| = ∞. So e xRln1 x dx diverges.
We
have
used
u
= ln x and
R 1
1
1
du = x dx to integrate x ln x dx = u du = ln |u| + C = ln | ln x| + C.
MATH 1760 : page 1 of 2
HW 7 Solution
MATH 1760 page 2 of 2
R∞
(c) e x(ln1x)2 dx Solution: We have used u = ln x and du = x1 dx to inteR
R
R∞
grate x(ln1x)2 dx = u12 du = − u1 + C = − ln1x + C and e x(ln1x)2 dx =
1
1 b
b→∞ − ln x |e = limb→∞ − ln b + 1 = 0 + 1 = 1.
Rlim
∞
(d) e lnxR2x dx Integration by partsR u = ln x, dv = Rx12 dx, du = x1 dx and
vR = x12 dx = − x1 + C , weR get lnx2x dx = − lnxx − (− x1 ) · x1 dx = − lnxx +
∞
1
dx = − lnxx − x1 +C. So e lnx2x dx = limb→∞ − lnxx − x1 |be limb→∞ − lnb b − 1b −
x2
(− lnee − 1e ) = 2e . We have used ln e = 1, limb→∞ 1b = 0 and limb→∞ lnb b =
0
1
limb→∞ (lnb0b) = limb→∞ 1b = limb→∞ 1b = 0
R∞
R∞
Rb
(e) −∞ x5 dx Since 0 x5 dx = limb→∞ 0 x5 dx limb→∞
diverges.
b6
6
= ∞, so
R∞
−∞
x5 dx