Solution to Quiz 7
• (50 pts)
R
4x3 −8x2 +2x−3
dx
x2 (x2 +1)
3
2
+2x−3
Solution: From the partial fraction 4x x−8x
= Ax + xB2 + Cx+D
, we get
2 (x2 +1)
x2 +1
3
2
2
2
2
3
2
4x −8x +2x−3 = Ax(x +1)+B(x +1)+(Cx+D)x , x −2x −2x+3 = Ax3 +
Ax+Bx2 +B+Cx3 +Dx2 and 4x3 −8x2 +2x−3 = (A+C)x3 +(B+D)x2 +Ax+B.
Thus A + C = 4, B + D = −8, A = 2 and B = −3 HenceA = 2, B = −3,
3
2 +2x−3
C = 4 − A = 2 and D = −8 − B = −5. Thus 4x x−8x
= x2 − x32 + 2x−5
2 (x2 +1)
x2 +1
R 4x3 −8x2 +2x−3
3
2
and
dx = 2 ln |x| + x + ln |x + 1| − 5 arctan(x) + C. We have
x2 (x2 +1) R
R 2x−5
used x2 +1 dx = x22x+1 dx − x25+1 dx = ln |x2 + 1| − 5 arctan(x) + C.
• (50 pts) Determine whether each integral is convergent or divergent.
If the integral is convergent, compute its value.
1.
Z ∞
1
1 dx
x3
1
b
R∞ 1
Rb 1
2
3 32 Solution: 1 1 dx = limb→∞ 1 1 dx = limb→∞ 2 x = limb→∞ 32 b 3 −
3
3
x
1
R∞x
3
= ∞. So 1 11 dx diverges.
2
x3
2.
Z
∞
1
dx
x(ln x)3
e
R
R
l We use u = ln x and du = x1 dx to integrate x(ln1x)3 dx = u13 du =
R∞
− 21 u−2 + C = − 2(ln1x)2 + C and e x(ln1x)3 dx = limb→∞ − 2(ln1x)2 |be =
limb→∞ − 2(ln1b)2 + 12 = 12 . We have used ln e = 1.
• (Bonus problem 20 pts)
Z
6x − 5
dx
+ 4x + 13
Solution: Completing the square, we get x2 + 4x + 13 = x2 + 4x + 4 +
9 = (x + 2)2 + 32 . Let x + 2 = 3u. Then dx = 3du and x = 3u − 2.
R 6·(3u−2)−5
R 18u−17
R 6x−5
R 6x−5
R 18u−17
dx
=
dx
=
3du
=
3du
=
du =
2
2
3
2
2
2
3 u+3
9(u +1)
3(u2 +1)
R x 6u+4x+13 17 R 1(x+2) +3
17
x+2 2
2
du − 3 u2 +1 du = 3 ln |u + 1| − 3 arctan(u) + C = 3 ln |( 3 ) + 1| −
u2 +1
17
arctan( x+2
) + C. In the last step, we have used u = x+2
.
3
3
3
x2
MATH 1760 : page 1 of 1