Chapter 6: Techniques of Integration
Section 6.5: Integrating the Fractions in a Partial-Fraction Decomposition
The indefinite integral of a rational function can be found if it is first subjected to a partial-fraction decomposition. At worst, the rational function ux/vx can be expressed as qx+rx/vx, where qx (the quotient) is a polynomial that yields to integration by the power rule, and rx is the remainder. The fraction r/v decomposes to fractions as per Table 6.4.1 in Section 6.4. Table 6.5.1 details the integration of the various partial fractions that the decomposition of r/v generates.
Integrates to a logarithm if m=1
Yields to the power rule if m>1
a x+bx2+B x+Cn
Complete the square in the denominator.
Set u=x2+B x +C so du=2 x+B dx. Then
∫a x+b dxx2+B x+Cn= ∫a/2 duun+∫b−a B/2 dxx+B/22+C−B2/4n
Set z=x+B/2,λ=b−a B/2, and σ2=C−B2/4. Then
∫a x+b dxx2+B x+Cn= ∫a/2 duun+∫λ dzz2±σ2n
The last integral yields to the trig substitutions in Table 6.3.1.
Table 6.5.1 Integrating partial fractions
Although the examples that follow will illustrate the machinations sketched in Table 6.5.1, this author believes that in this day and age, any reasonable consumer of mathematics will implement the integral of a rational function either with software such as Maple, or at the very least, by the expediency of a table of integrals. This author's table of integrals, in use for more than 50 years, provides the excerpts shown in Table 6.5.2.
∫dxQ=2qtan−12 a x+bq
∫dxQ=−2−qtanh−12 a x+b−q
∫dxQ=1−qln2 a x+b−−q2 a x+b+−q
∫dxQn+1=2 a x+bn q Qn+2 a 2 n−1n q∫dxQn
∫x dxQ=12 alnQ−b2 a∫dxQ
∫x dxQn+1=−2 c+b xn q Q−b 2 n−1n q∫dxQn
Table 6.5.2 Excerpts from a table of integrals
Q=a x2+b x+c and q=4 a c−b2
The first three formulas in Table 6.5.2 would be used for the integral of a constant divided by a quadratic; the fourth, for the integral of a constant divided by a power of a quadratic. The fifth formula in Table 6.5.2 would be used for the integral of x divided by a quadratic; the sixth, for the integral of x divided by a power of a quadratic. Hence, the fourth and sixth formulas in the table are actually reduction formulas.
Evaluate the integral ∫7⁢x−23x2−7 x+12 ⅆx.
Evaluate the integral ∫7⁢x3−86⁢x2+346⁢x−455x4−15⁢x3+84⁢x2−208 x+192 ⅆx.
Evaluate the integral ∫5⁢x3−11⁢x2+18⁢x+1x4−5⁢x3+14⁢x2−19⁢x+15ⅆx.
Evaluate the integral ∫5⁢x4+37⁢x3+15⁢x2−150⁢x+109x5+11⁢x4+21⁢x3−59⁢x2−21 x+49 ⅆx.
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