Seja Q(x)=(x2+ 1)4⇒Q0(x)=8x(x2+ 1)3;Q1(x) = (x2+ 1)3eQ2(x) = x2+ 1
Zdx
(x2+ 1)4=Ax5+Bx4+C x3+Dx2+Ex +F
(x2+ 1)3+ZGx +H
x2+ 1 dx
Derivando ambos os membros da igualdade, teremos:
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(x2+ 1)4=(5Ax4+ 4Bx3+ 3C x2+ 2Dx +E)·(x2+ 1)3
(x2+ 1)6−
−(Ax5+Bx4+C x3+Dx2+Ex +F)·(x2+ 1)2·6x
(x2+ 1)6+(Gx +H)
x2+ 1 , ent˜ao:
x4+2x2+ 1 = Gx11 + (−A+H)x10 + (−2B+ 5G)x9+ (3A−3C+ 5H)x8+ (−4D+ 10G)x7+
+(9A−3C−5E+ 10H)x6+ (6B−6D−6F+ 10G)x5+ (5A+ 3C−9E+ 10H)x4+
+(4B−12F+ 5G)x3+ (3C−3E+ 5H)x2+ (2D−6F+G)x+ (E+H)
E+H= 1
2D−6F+G= 0
3C−3E+ 5H= 2
4B−12F+ 5G= 0
5A+ 3C−9E+ 10H= 1
6B−6D−6F+ 10G= 0
9A−3C−5E+ 10H= 0
−4D+ 10G= 0
3A−3C+ 5H= 0
−2B+ 5G= 0
−A+H= 0
G= 0
, resolvendo o sistema, achamos:
A=5
16;B= 0; C=5
6;D= 0; E=11
16;F= 0; G= 0; H=5
16, logo:
Zdx
(x2+ 1)4=Ax5+Bx4+C x3+Dx2+Ex +F
(x2+ 1)3+ZGx +H
x2+ 1 dx =
=
5
16x5+5
6x3+11
16x
(x2+ 1)3+5
16 Zdx
x2+ 1 =15x5+ 40x3+ 33x
48(x2+ 1)3+15
48 arctan x+C.
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