Chapter 6: Techniques of Integration
Section 6.5: Integrating the Fractions in a Partial-Fraction Decomposition
Evaluate the integral ∫5⁢x3−11⁢x2+18⁢x+1x4−5⁢x3+14⁢x2−19⁢x+15ⅆx.
From the partial-fraction decomposition in Example 6.4.4, it follows that
∫5⁢x3−11⁢x2+18⁢x+1x4−5⁢x3+14⁢x2−19⁢x+15ⅆx= ∫3⁢x+2x2−3⁢x+5ⅆx+∫2⁢x−1x2−2 x+3ⅆx
Apply the ideas of Table 6.5.1 to the first integral on the right, to obtain
where z=x−3/2 and σ=11/2. From Table 6.3.1, the substitution z=σ tanθ changes the second integral on the right to
Apply these same ideas to the second integral arising from the partial-fraction decomposition, to obtain
where z=x−1 and σ=2. From Table 6.3.1, the substitution z=σ tanθ in the second integral on the right becomes
∫dzz2+σ2 = 1σ∫dθ = 12 arctanθ=12 arctanx−12
Consequently, the value of the given integral is
32lnx2−3 x+5 +1311arctan2 x−311 +lnx2−2 x+3 +12 arctanx−12
where the logarithms arise from the integrals of du/u, and absolute values are not needed because the quadratic arguments are strictly positive.
Solution via Table of Integrals
Let Q1=x2−3 x+5 and Q2=x2−2 x+3 so in the notation of Table 6.5.2, a1=a2=1, b1=−3,b2=−2, c1=5,c2=3, and q1=11,q2=8. Noting formulas (1) and (5) in Table 6.5.2, write
=312 a1lnQ1−b12 a1∫dxQ1+2∫dxQ1
=212 a2lnQ2−b22 a2∫dxQ2−∫dxQ2
Combining these results leads to
32lnx2−3 x+5+lnx2−2 x+3+1311arctan2 x−311+12arctanx−12
as the value of the given indefinite integral.
Evaluation in Maple
Control-drag the given integral.
Context Panel: Evaluate and Display Inline
∫5⁢x3−11⁢x2+18⁢x+1x4−5⁢x3+14⁢x2−19⁢x+15ⅆx = ln⁡x2−2⁢x+3+12⁢2⁢arctan⁡14⁢2⁢x−2⁢2+32⁢ln⁡x2−3⁢x+5+1311⁢11⁢arctan⁡111⁢2⁢x−3⁢11
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.
Table 6.5.3(a) shows the result of invoking the Partial Fractions rule in the
tutor when the Sum rule is taken as an Understood Rule.
Table 6.5.3(a) Partial Fractions rule applied in Integration Methods tutor
An annotated stepwise solution for the integration of the first partial fraction appears in Table 6.5.3(b). It is generated by the
tutor with the Constant, Constant Multiple, and Sum rules taken as Understood Rules. Algebra is used to split the integral to a du/u term and a dx/Q1 term. Rather than apply completion of the square to this latter term, Maple makes the substitution u=x−3/2, then makes the trig substitution in terms of the tangent function. Unfortunately, Maple insists on rationalizing the reciprocal of a square root so that 11/11 is seen, instead of the more compact 1/11.
Table 6.5.3(b) Integration of the first partial fraction via the Integration Methods tutor
An annotated stepwise solution for the integration of the second partial fraction appears in Table 6.5.3(c). It is generated by the
tutor with the Constant, Constant Multiple, and Sum rules taken as Understood Rules. Algebra is used to split the integral to a du/u term and a dx/Q2 term. Rather than apply completion of the square to this latter term, Maple makes the substitution u=x−1, then makes the trig substitution in terms of the tangent function. Unfortunately, Maple insists on rationalizing the reciprocal of a square root so that 2/2 is seen, instead of the more compact 1/2.
Table 6.5.3(c) Integration of the second partial fraction via the Integration Methods tutor
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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