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Calculus II, Section 7.4, #28
Integration of Rational Functions by Partial Fractions
Evaluate the integral.1
Z 3
x + 6x − 2
dx
x4 + 6x2
Is the integrand one of our basic indefinite integrals? No. How about a basic u-substitution? No. Integration
by parts? No. Powers of trig functions? No. Does the integrand include a trig. sub. radical? No. Is the
integrand a rational function? YES!
The degree of the numerator is 3, and the degree of the denominator is 4. Since the degree of the numerator
is less than the degree of the denominator, we are ready to begin the partial fractions process.
Consider the integrand
x3 + 6x − 2
x3 + 6x − 2
=
x4 + 6x2
x2 (x2 + 6)
x2 is a repeated linear factor, whereas x2 + 6 is a distinct, irreducible quadratic factor. We have
The LCD is x2
x3 + 6x − 2
A
B
Cx + D
= + 2+ 2
x2 (x2 + 6)
x
x
x +6
x2 + 6 , and we multiply both sides of this identity by the LCD to get
A
B
Cx + D 2 2
x3 + 6x − 2 2 2
· x x + 6 = · x2 x2 + 6 + 2 · x2 x2 + 6 + 2
·x x +6
x2 (x2 + 6)
x
x
x +6
x3 + 6x − 2 = Ax x2 + 6 + B x2 + 6 + (Cx + D) x2
x3 + 6x − 2 = A x3 + 6x + B x2 + 6 + (Cx + D) x2
x3 + 6x − 2 = Ax3 + 6Ax + Bx2 + 6B + Cx3 + Dx2
x3 + 6x − 2 = Ax3 + Cx3 + Bx2 + Dx2 + 6Ax + 6B
x3 + 6x − 2 = (A + C) x3 + (B + D) x2 + 6Ax + 6B
Since this equation is an identity, i.e., it is true for all allowable values of x, we equate coefficients to get the
system of equations

1 =
A
+ C



0 =
B +
+ D
6
=
6A



−2 =
6B
From the third equation we get A = 1. From the fourth we get B = − 13 . Substituting our value for A into the
first equation gives us C = 0. Substituting our value for B into the second equation gives us D = 31 .
The partial fraction decomposition is
x3 + 6x − 2
1 −1/3 0 · x + 1/3
= +
+
2
2
x (x + 6)
x
x2
x2 + 6
1 −1/3
1/3
= +
+ 2
x
x2
x +6
1 Stewart,
Calculus, Early Transcendentals, p. 501, #28.
Calculus II
Integration of Rational Functions by Partial Fractions
so our integral becomes
Z 3
Z
x + 6x − 2
dx
=
x4 + 6x2
Z
=
Z
=
Z
=
1
x
1
x
1
x
1
x
1/3
−1/3
+ 2
dx
2
x
x +6
Z
Z
1/3
−1/3
dx +
dx +
dx
x2
x2 + 6
Z
Z
1
1
1
1
dx +
dx −
dx
2
2
3
x
3
x +6
Z
Z
1
1
1
x−2 dx +
dx −
√ 2 dx
3
3
2
x +
6
+
The first integral is a straightforward natural logarithm, the second integral is a straightforward power rule,
and the third integral is an inverse tangent form.
1
1
1 1
x
= ln |x| −
−
+ · √ tan−1 √
+C
3
x
3
6
6
1
1
x
= ln |x| +
+ √ tan−1 √
+C
3x 3 6
6
Thus,
Z
x3 + 6x − 2
1
1
dx = ln |x| +
+ √ tan−1
x4 + 6x2
3x 3 6
x
√
6
+C