Chapter 6: Techniques of Integration
Section 6.5: Integrating the Fractions in a Partial-Fraction Decomposition
Evaluate the integral ∫5⁢x4+37⁢x3+15⁢x2−150⁢x+109x5+11⁢x4+21⁢x3−59⁢x2−21 x+49 ⅆx.
From the partial-fraction decomposition in Example 6.4.5, it follows that
∫5⁢x4+37⁢x3+15⁢x2−150⁢x+109x5+11⁢x4+21⁢x3−59⁢x2−21 x+49ⅆx= ∫2x+1ⅆx+∫3⁢x−1x2+5⁢x−7ⅆx+∫5⁢x+4x2+5⁢x−72ⅆx
The first integral on the right evaluates to 2 ln(x+1). Apply the ideas of Table 6.5.1 to the second integral on the right, to obtain
∫3⁢x−1x2+5 x−7ⅆx=32∫duu−172∫dzz2− σ2
where z=x+5/2 and σ=53/2. From Table 6.3.1, the substitution z=σ secθ changes the second integral on the right to
=−172 σ∫cscθ dθ
=−17253ln(2 x+5−532 x+5+53)
Apply these same ideas to the third integral arising from the partial-fraction decomposition, to obtain
= −172∫σ secθtanθσ2sec2θ−12 dθ
=−172 σ3∫secθtan3θ dθ
=−172 σ3∫cos2θsin3θ dθ
=−172 σ3∫1−sin2θsin3θ dθ
=−172 σ3∫csc3θ dθ−∫cscθ dθ
= 174 σ3(cscθcotθ+lncscθ−cotθ)
= 174 σ3(σ zz2−σ2+12ln(z−σz+σ))
=171062 x+5x2+5 x−7+175353ln(2 x+5−532 x+5+53)
where the integral of csc3θ is detailed in Example 6.2.10.
Consequently, the value of the given integral is the sum of the two computed integrals, the log term, and −5/2x2+5 x−7, that is,
This expression is real for all real x, except for x=−1, −5+53/2≐−6.14, and −5−53/2≐1.14, where both the integrand and the integral have vertical asymptotes.
Solution via Table of Integrals
Let Q=x2+5 x−7, so in the notation of Table 6.5.2, a=1, b=5, c=−7, n=1, and q=−53. Noting formulas (3) and (5) in Table 6.5.2, write
=312 alnQ−b2 a∫dxQ−∫dxQ
=32lnQ−172153ln2 x+5−532 x+5+53
Noting formulas (3) - (6) in Table 6.5.2, write
∫5 x+4Q2 dx
=5−b x+2 cn q Q−b2 n−1n q∫dxQ+42 a x+bn q Q+2 a2 n−1n q∫dxQ
=55 x−1453 Q+553∫dxQ+42 x+5−53 Q−253∫dxQ
=55 x−14−42 x+553 Q+1753∫dxQ
=17 x−9053 Q+175353ln2 x+5−532 x+5+53
Combining these results leads to
32lnQ−86710653ln2 x+5−532 x+5+53+17 x−9053 Q+2 lnx+1
as the value of the given indefinite integral.
Evaluation in Maple
Control-drag the given integral and press the Enter key.
Note once again that Maple integrates 1/x to lnx, not ln(x), relying on a complex constant of integration to counterbalance the logarithm of a negative number. That is probably irrelevant, as once again Maple has returned an antiderivative that is not real for any value of real x other than x=−1, and x=−5 ±53/2 where there are vertical asymptotes.
The partial-fraction decomposition of the integrand, as computed by Maple's Context Panel system, is obtained in Table 6.5.4(a) where the underlying command is convert/parfrac.
Control-drag the integrand.
Context Panel: Conversions≻Partial Fractions≻x
5⁢x4+37⁢x3+15⁢x2−150⁢x+109x5+11⁢x4+21⁢x3−59⁢x2−21 x+49 → 2x+1+3⁢x−1x2+5⁢x−7+5⁢x+4x2+5⁢x−72
Table 6.5.4(a) Partial-fraction decomposition of the integrand.
The Partial Fractions rule in the Integration Methods tutor factors the quadratic in the denominators of the partial fractions in Table 6.5.4(a) to linear factors, making more terms, and terms of greater visual complexity because the factors are x−−5 ±53/2. Table 6.5.4(b) summarizes the five terms that the
tutor utilizes for the partial-fraction decomposition and ensuing integration.
Table 6.5.4(b) Partial fractions and integration as per the Integration Methods tutor
Unfortunately, the sum of these five terms is real only for x∈−1,1. If the arguments of the logarithmic terms were wrapped in absolute-value bars, then the solution generated by the tutor would have a greater domain.
Note that an annotated stepwise solution is available via the Context Panel with the "All Solution Steps" option.
The rules of integration can also be applied via the Context Panel, as per the figure to the right.
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